The document discusses the key differences between combinations and permutations. A combination is an unordered selection of objects, while a permutation is an ordered selection where the order matters. Some key points made include: - A combination does not consider the order of selected objects, while a permutation does. - The number of permutations of n objects taken r at a time is written as nPr and is calculated as n!/(n-r)!, while the number of combinations is written as nCr and is calculated as nPr/r!. - Examples are given to illustrate the differences, such as forming a number using digits versus selecting players for a team where order does not matter. - Formulas are provided

1. Combinations and Permutations
What's the Difference?
In English we use the word "combination" loosely, without
thinking if the order of things is important. In other words:
"My fruit salad is a combination of apples, grapes and
bananas" We don't care what order the fruits are in,
they could also be "bananas, grapes and apples" or
"grapes, apples and bananas", its the same fruit salad.
"The combination to the safe was 472". Now we do
care about the order. "724" would not work, nor would
"247". It has to be exactly 4-7-2.
So, in Mathematics we use more precise language:
If the order doesn't matter, it is a Combination.
If the order does matter it is a Permutation.
A Permutation is an ordered Combination.
Permutation : Permutation means arrangement of things. The
word arrangement is used, if the order of things is considered.
Combination: Combination means selection of things. The word
selection is used, when the order of things has no importance.
Definition:
Permutation:
An arrangement is called a Permutation. It is the
rearrangement of objects or symbols into distinguishable sequences.
When we set things in order, we say we have made an arrangement.
When we change the order, we say we have changed the

2. arrangement. So each of the arrangement that can be made by
taking some or all of a number of things is known as Permutation.
Combination:
A Combination is a selection of some or all of a number of
different objects. It is an un-ordered collection of unique sizes.In a
permutation the order of occurence of the objects or the
arrangement is important but in combination the order of occurence
of the objects is not important.
Formula:
Permutation = nPr = n! / (n-r)!
Combination = nCr = nPr / r!
where,
n, r are non negative integers and 0≤ r≤n.
r is the size of each permutation.
n is the size of the set from which elements are permuted.
! is the factorial operator.
Example: Suppose we have to form a number of consisting
of three digits using the digits 1,2,3,4, To form this number the
digits have to be arranged. Different numbers will get formed
depending upon the order in which we arrange the digits. This
is an example of Permutation.
Now suppose that we have to make a team of 11 players out of
20 players, This is an example of combination, because the
order of players in the team will not result in a change in the
team. No matter in which order we list out the players the team
will remain the same! For a different team to be formed at least
one player will have to be changed.
Now let us look at two fundamental principles of counting:

3. Addition rule : If an experiment can be performed in „n‟ ways, &
another experiment can be performed in „m‟ ways then either of
the two experiments can be performed in (m+n) ways. This rule
can be extended to any finite number of experiments.
Example: Suppose there are 3 doors in a room, 2 on one
side and 1 on other side. A man wants to go out from the room.
Obviously he has „3‟ options for it. He can come out by door „A‟
or door „B‟ or door ‟C‟.
Multiplication Rule : If a work can be done in m ways, another
work can be done in „n‟ ways, then both of the operations can
be performed in m x n ways. It can be extended to any finite
number of operations.
Example.: Suppose a man wants to cross-out a room, which
has 2 doors on one side and 1 door on other site. He has 2 x
1 = 2 ways for it.
Factorial n : The product of first „n‟ natural numbers is denoted
by n!.
n! = n(n-1) (n-2) ………………..3.2.1.
Ex. 5! = 5 x 4 x 3 x 2 x 1 =120
Note 0! = 1

4. Proof n! =n, (n-1)!
Or (n-1)! = [n x (n-1)!]/n = n! /n
Putting n = 1, we have
O! = 1!/1 = 1
Permutation
Number of permutations of „n‟ different things taken „r‟ at a
time is given by:-
n
Pr = n!/(n-r)!
Proof: Say we have „n‟ different things a1, a2……, an.
Clearly the first place can be filled up in „n‟ ways. Number of
things left after filling-up the first place = n-1
So the second-place can be filled-up in (n-1) ways. Now
number of things left after filling-up the first and second
places = n - 2
Now the third place can be filled-up in (n-2) ways.
Thus number of ways of filling-up first-place = n
Number of ways of filling-up second-place = n-1
Number of ways of filling-up third-place = n-2
Number of ways of filling-up r-th place = n – (r-1) = n-r+1
By multiplication – rule of counting, total no. of ways of filling
up, first, second -- rth-place together :-

5. n (n-1) (n-2) ------------ (n-r+1)
Hence:
n
Pr = n (n-1)(n-2) --------------(n-r+1)
= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-
1)] ----3.2.1
n
Pr = n!/(n-r)!
Number of permutations of „n‟ different things taken all at a
time is given by:-
n
Pn = n!
Proof :
Now we have „n‟ objects, and n-places.
Number of ways of filling-up first-place = n
Number of ways of filling-up second-place = n-1
Number of ways of filling-up third-place = n-2
Number of ways of filling-up r-th place, i.e. last place =1
Number of ways of filling-up first, second, --- n th place
= n (n-1) (n-2) ------ 2.1.
n
Pn = n!
Concept.
n
We have Pr = n!/n-r
Putting r = n, we have :-

6. n
Pr = n! / (n-r)
n
But Pn = n!
Clearly it is possible, only when n! = 1
Hence it is proof that 0! = 1
Note : Factorial of negative-number is not defined. The
expression –3! has no meaning.
Examples
Q. How many different signals can be made by 5 flags from 8-
flags of different colours?
Ans. Number of ways taking 5 flags out of 8-flage = 8P5
= 8!/(8-5)!
= 8 x 7 x 6 x 5 x 4 = 6720
Q. How many words can be made by using the letters of the
word “SIMPLETON” taken all at a time?
Ans. There are „9‟ different letters of the word “SIMPLETON”
Number of Permutations taking all the letters at a time = 9P9
= 9! = 362880.
Number of permutations of n-thing, taken all at a time, in which
„P‟ are of one type, „q‟ of them are of second-type, „r‟ of
them are of third-type, and rest are all different is given by
:-
n!/(p! x q! x
r!)

7. Example: In how many ways can the letters of the word “Pre-
University” be arranged?
13!/(2! X 2! X 2!)
Number of permutations of n-things, taken „r‟ at a time when
each thing can be repeated r-times is given by = nr.
Proof.
Number of ways of filling-up first –place = n
Since repetition is allowed, so
Number of ways of filling-up second-place = n
Number of ways of filling-up third-place
Number of ways of filling-up r-th place = n
Hence total number of ways in which first, second ----r th,
places can be filled-up
= n x n x n ------------- r factors.
= nr
Example: John has 8 friends. In how many ways can he invite
one or more of them to dinner?
Ans. John can select one or more than one of his 8 friends.
=> Required number of ways = 28 – 1= 255.

8. (iv) Number of ways of selecting zero or more things from „n‟
identical things is given by :- n+1
Example: In how many ways, can zero or more letters be
selected form the letters AAAAA?
Ans. Number of ways of :
Selecting zero 'A's = 1
Selecting one 'A's = 1
Selecting two 'A's =1
Selecting three 'A's = 1
Selecting four 'A's = 1
Selecting five 'A's = 1
=> Required number of ways = 6 [5+1]
(V) Number of ways of selecting one or more things from „p‟
identical things of one type „q‟ identical things of another
type, „r‟ identical things of the third type and „n‟ different
things is given by :-
(p+1) (q+1) (r+1)2n – 1

9. Example: Find the number of different choices that can be
made from 3 apples, 4 bananas and 5 mangoes, if at least
one fruit is to be chosen.
Ans:
Number of ways of selecting apples = (3+1) = 4 ways.
Number of ways of selecting bananas = (4+1) = 5 ways.
Number of ways of selecting mangoes = (5+1) = 6 ways.
Total number of ways of selecting fruits = 4 x 5 x 6
But this includes, when no fruits i.e. zero fruits is selected
=> Number of ways of selecting at least one fruit = (4x5x6) -1 =
119
Note :- There was no fruit of a different type, hence here n=o
=> 2n = 20=1
(VI) Number of ways of selecting „r‟ things from „n‟ identical
things is „1‟.
Example: In how many ways 5 balls can be selected from „12‟
identical red balls?
Ans. The balls are identical, total number of ways of selecting 5
balls = 1.
Example: How many numbers of four digits can be formed with
digits 1, 2, 3, 4 and 5?

10. Ans. Here n = 5 [Number of digits]
And r = 4 [ Number of places to be filled-up]
5
Required number is P4 = 5!/1! = 5 x 4 x 3 x 2 x 1
Example: A child has 3 pocket and 4 coins. In how many
ways can he put the coins in his pocket.
Ans. First coin can be put in 3 ways, similarly second, third
and forth coins also can be put in 3 ways.
So total number of ways = 3 x 3 x 3 x 3 = 34 = 81
So, we should really call this a "Permutation
Lock"!
Permutations
There are basically two types of permutation:
1. Repetition is Allowed: such as the lock above. It could be
"333".
2. No Repetition: for example the first three people in a running
race. You can't be first and second.

11. 1. Permutations with Repetition
These are the easiest to calculate.
When you have n things to choose from ... you have n choices each
time!
When choosing r of them, the permutations are:
n × n × ... (r times)
(In other words, there are n possibilities for the first choice, THEN
there are n possibilites for the second choice, and so on, multplying
each time.)
Which is easier to write down using an exponent of r:
n × n × ... (r times) = nr
Example: in the lock above, there are 10 numbers to choose from
(0,1,..9) and you choose 3 of them:
10 × 10 × ... (3 times) = 103 = 1,000 permutations
So, the formula is simply:
nr
where n is the number of
things to choose from, and
you choose r of them
(Repetition allowed, order
matters)

12. 2. Permutations without Repetition
In this case, you have to reduce the number of available choices each
time.
For example, what order could 16
pool balls be in?
After choosing, say, number "14"
you can't choose it again.
So, your first choice would have 16 possibilites, and your next choice
would then have 15 possibilities, then 14, 13, etc. And the total
permutations would be:
16 × 15 × 14 × 13 × ... = 20,922,789,888,000
But maybe you don't want to choose them all, just 3 of them, so that
would be only:
16 × 15 × 14 = 3,360
In other words, there are 3,360 different ways that 3 pool balls could
be selected out of 16 balls.
But how do we write that mathematically? Answer: we use the
"factorial function"
The factorial function (symbol: !) just means
to multiply a series of descending natural
numbers. Examples:
4! = 4 × 3 × 2 × 1 = 24
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040
1! = 1
Note: it is generally agreed that 0! = 1. It may seem
funny that multiplying no numbers together gets you 1,

13. but it helps simplify a lot of equations.
So, if you wanted to select all of the billiard balls the permutations
would be:
16! = 20,922,789,888,000
But if you wanted to select just 3, then you have to stop the
multiplying after 14. How do you do that? There is a neat trick ... you
divide by 13! ...
16 × 15 × 14 × 13 × 12 ...
= 16 × 15 × 14 = 3,360
13 × 12 ...
Do you see? 16! / 13! = 16 × 15 × 14
The formula is written:
where n is the number of
things to choose from, and
you choose r of them
(No repetition, order matters)
Examples:
Our "order of 3 out of 16 pool balls example" would be:
16! 16! 20,922,789,888,000
= = = 3,360
(16-3)! 13! 6,227,020,800
(which is just the same as: 16 × 15 × 14 = 3,360)

14. How many ways can first and second place be awarded to 10 people?
10! 10! 3,628,800
= = = 90
(10-2)! 8! 40,320
(which is just the same as: 10 × 9 = 90)
Notation
Instead of writing the whole formula, people use different notations
such as these:
Example: P(10,2) = 90
Combinations
There are also two types of combinations (remember the order does
not matter now):
1. Repetition is Allowed: such as coins in your pocket
(5,5,5,10,10)
2. No Repetition: such as lottery numbers (2,14,15,27,30,33)
How many permutations of 3 different digits are there, chosen from
the ten digits 0 to 9 inclusive?
A
84
B
120
C

15. 504
D
720
Answer :The number of permutations of 3 digits chosen from 10 is
10
P3 = 10 × 9 × 8 = 720
Jones is the Chairman of a committee. In how many ways can a
committee of 5 be chosen from 10 people given that Jones must be
one of them?
A
126
B
252
C
495
D
3,024
Answer: Jones is already chosen, so we need to choose another 4
from 9.
In choosing a committee, order doesn't matter; so we need the
number of combinations of 4 people chosen from 9
= 9 C4
= 9!/(4!)(5!)
= (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1)
= 3,024/24 = 126
A password consists of four different letters of the alphabet. How
many different possible passwords are there?
(Using pen and paper can help you learn.)
A
426

16. B
456,976
C
14,950
D
358,800
Answer: The number of permutations of 4 letters chosen from 26 is
26
P4 = 26 × 25 × 24 × 23 = 358,800
A password consists of two letters of the alphabet followed by three
digits chosen from 0 to 9. Repeats are allowed. How many different
possible passwords are there?
(Don't worry if you get it wrong ... you can learn from
your mistakes.)
A
492,804
B
650,000
C
676,000
D
1,757,600
Answer: The number of ways of choosing the letters = 26 × 26 = 676
The number of ways of choosing the digits = 10 × 10 × 10 = 1,000
So the number of possible passwords = 676 × 1,000 = 676,000
An encyclopedia has eight volumes. In how many ways can the eight
volumes be replaced on the shelf?
(Use pen and paper to work out the answer.)
A
8
B

17. 5,040
C
40,320
D
88
Answer:Imagine there are 8 spots on the shelf. Replace the volumes
one by one.
The first volume to be replaced could go in any one of the eight spots.
The second volume to be replaced could then go in any one of the
seven remaining spots.
The third volume to be replaced could then go in any one of the six
remaining spots.
etc
So the total number of ways the eight volumes could be replaced
= 8!
=8×7×6×5×4×3×2×1
= 40,320
Assuming that any arrangement of letters forms a 'word', how many
'words' of any length can be formed from the letters of the word
SQUARE?
(No repeating of letters)
(No guessing! Be sure of your answer.)
A
82
B
720
C
1,956
D
9,331
Answer: The number of one letter 'words' = 6P1 = 6
The number of two letter 'words' = 6P2 = 6 × 5 = 30

18. The number of three letter 'words' = 6P3 = 6 × 5 × 4 = 120
The number of four letter 'words' = 6P4 = 6 × 5 × 4 × 3 = 360
The number of five letter 'words' = 6P5 = 6 × 5 × 4 × 3 × 2 = 720
The number of six letter 'words' = 6P6 = 6! = 720
So the total number of possible 'words' = 6 + 30 + 120 + 360 + 720 +
720 = 1,956
16 teams enter a competition. They are divided up into four Pools (A,
B, C and D) of four teams each.
Every team plays one match against the other teams in its Pool.
After the Pool matches are completed:
• the winner of Pool A plays the second placed team of Pool B
• the winner of Pool B plays the second placed team of Pool A
• the winner of Pool C plays the second placed team of Pool D
• the winner of Pool D plays the second placed team of Pool C
The winners of these four matches then play semi-finals, and the
winners of the semi-finals play in the final.
How many matches are played altogether?
(Using pen and paper can help you learn.)
A
23
B
31
C
32
D
63
Answer:The number of matches played in each Pool = 4C2 = 4!/(2!2!)
= (4 × 3)/(2 × 1) = 6

19. So the total number of Pool matches = 4 × 6 = 24
The winners and second placed teams play a further 4 matches.
Then there are 2 semi-finals and 1 final
So the total number of matches = 24 + 4 + 2 + 1 = 31
A restaurant offers 5 choices of appetizer, 10 choices of main meal
and 4 choices of dessert. A customer can choose to eat just one
course, or two different courses, or all three courses. Assuming all
choices are available, how many different possible meals does the
restaurant offer?
A
329
B
310
C
200
D
19
Answer; A person who eats only an appetizer has 5 choices.
A person who eats only a main meal has 10 choices.
A person who eats only a dessert has 4 choices.
A person who eats an appetizer and a main meal has 5 × 10 = 50
choices.
A person who eats an appetizer and a dessert has 5 × 4 = 20 choices.
A person who eats a main meal and a dessert has 10 × 4 = 40 choices.
A person who eats all three courses has 5 × 10 × 4 = 200 choices
So the total number of possible meals = 5 + 10 + 4 + 50 + 20 + 40 +
200 = 329

20. Question:
How many ways can 4 prizes be given away to 3 boys, if each
boy is eligible for all the prizes?
(1)256
(2)12
(3)81
(4) None of these
Correct Answer - (3)
Solution:
Any one prize can be given to any one of the 3 boys and hence
there are 3 ways of distributing each prize.
Hence, the 4 prizes can be distributed in 34= 81 ways.
Question:
How many words of 4 consonants and 3 vowels can be made
from 12 consonants and 4 vowels, if all the letters are
different?
(1)16C7.7!
(2)12C4.4C3.7!

21. (3)12C3.4C4
(4) 12C4 . 4C3
Correct Answer - (2)
Solution:
12
4 consonants out of 12 can be selected in C4 ways.
3 vowels can be selected in 4C3 ways.
Therefore, total number of groups each containing 4
consonants and 3 vowels = 12C4 . 4C3
Each group contains 7 letters, which can be arranging in 7!
ways.
12
Therefore required number of words = 4 . 4C3 . 7!
Question:
In how many ways can the letters of the word EDUCATION be
rearranged so that the relative position of the vowels and
consonants remain the same as in the word EDUCATION?
(1)9!/4
(2)9!/(4!.5!)
(3)4!.5!
(4) None of these
Correct Answer - (3)

22. Solution:
The word EDUCATION is a 9 letter word, with none of the
letters repeating.
The vowels occupy 3, 5, 7th and 8th position in the word and
the remaining 5 positions are occupied by consonants
As the relative position of the vowels and consonants in any
arrangement should remain the same as in the word
EDUCATION, the vowels can occupy only the aforementioned 4
places and the consonants can occupy 1st, 2nd, 4th, 6th and 9th
positions.
The 4 vowels can be arranged in the 3rd, 5th, 7th and 8th
position in 4! Ways.
Similarly, the 5 consonants can be arranged in 1st, 2nd, 4th, 6th
and 9th position in 5! Ways.
Hence, the total number of ways = 4! . 5!.
Question:
There are 12 yes or no questions. How many ways can these
be answered?
(1)1024
(2)2048
(3) 4096
(4) 144
Correct Answer - (3)

23. Solution:
Each of the questions can be answered in 2 ways (yes or no)
Therefore, no. of ways of answering 12 questions = 212 = 4096
ways.
Question:
What is the value of 1.1! + 2.2! + 3!.3! + ............ n.n!,
where n! means n factorial or n(n-1)(n-2)...1
Solution:
1.1! = (2 -1).1! = 2.1! – 1.1! = 2! - 1!
2.2! = (3 - 1).2! = 3.2! - 2! = 3! - 2!
3.3! = (4 - 1).3! = 4.3! - 3! = 4! - 3!
..
..
..
n.n! = (n+1 - 1).n! = (n+1)(n!) - n! = (n+1)! - n!
Summing up all these terms, we get (n+1)! - 1!
Question:
In how many ways can the letters of the word MANAGEMENT
be rearranged so that the two As do not appear together?
Solution:
The word MANAGEMENT is a 10 letter word.
Normally, any 10 letter word can be rearranged in 10! ways.
However, as there are certain letters of the word repeating,
we need to account for those. In this case, the letters A, M, E
and N repeat twice each.

24. Therefore, the number of ways in which the letters of the
word MANAGEMENT can be rearranged reduces to .
The problem requires us to find out the number of outcomes
in which the two As do not appear together.
The number of outcomes in which the two As appear together
can be found out by considering the two As as one single
letter. Therefore, there will now be only 9 letters of which
three of them E, N and M repeat twice. So these 9 letters with
3 of them repeating twice can be rearranged in ways.
Therefore, the required answer in which the two As do not
appear next to each other =
Total number of outcomes - the number of outcomes in which
the 2 As appear together
=> ways.
**Question:
There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs.
How many different albums can be formed using the above
repertoire if the albums should contain at least 1 Rock song
and 1 Carnatic song?
(1)15624
(2)16384
(3)6144
(4) 240

25. Correct Answer - (1)
Solution:
There are 2n ways of choosing ‘n’ objects. For e.g. if n = 3,
then the three objects can be chosen in the following 23 ways
- 3C0 ways of choosing none of the three, 3C1 ways of choosing
one out of the three, 3C2 ways of choosing two out of the
three and 3C3 ways of choosing all three.
In the given problem, there are 5 Rock songs. We can choose
them in 25 ways. However, as the problem states that the
case where you do not choose a Rock song does not exist (at
least one rock song has to be selected), it can be done in 25 -
1 = 32 - 1 = 31 ways.
Similarly, the 6 Carnatic songs, choosing at least one, can be
selected in 26 - 1 = 64 - 1 = 63 ways.
And the 3 Indi pop can be selected in 23 = 8 ways. Here the
option of not selecting even one Indi Pop is allowed.
Therefore, the total number of combinations = 31 . 63 . 8 =
15624
Question:
How many words can be formed by re-arranging the letters of
the word ASCENT such that A and T occupy the first and last
position respectively?
(1)5!
(2)4!

26. (3)2!
(4) 6! / 2!
Correct Answer - (2)
Solution:
As A and T should occupy the first and last position, the first
and last position can be filled in only one way. The remaining
4 positions can be filled in 4! Ways by the remaining words
(S,C,E,N,T). hence by rearranging the letters of the word
ASCENT we can form 1x4! = 4! words.
**Question:
A team of 8 students goes on an excursion, in two cars, of
which one can seat 5 and the other only 4. In how many ways
can they travel?
Solution:
There are 8 students and the maximum capacity of the cars
together is 9.
We may divide the 8 students as follows
Case I: 5 students in the first car and 3 in the second
Or Case II: 4 students in the first car and 4 in the second
Hence, in Case I: 8 students are divided into groups of 5 and 3
in 8C3 ways.
Similarly, in Case II: 8 students are divided into two groups of
4 and 4 in 8C4 ways.

27. Therefore, the total number of ways in which 8 students can
travel is 8C3 + 8C4 = 56 + 70 = 126.
**Question:
When four fair dice are rolled simultaneously, in how many
outcomes will at least one of the dice show 3?
Solution:
When 4 dice are rolled simultaneously, there will be a total of
64 = 1296 outcomes.
The number of outcomes in which none of the 4 dice show 3
will be 54 = 625 outcomes.
Therefore, the number of outcomes in which at least one die
will show 3 = 1296 – 625 = 671
Question:
How many ways can 10 letters be posted in 5 post boxes, if
each of the post boxes can take more than 10 letters?
Solution:
Each of the 10 letters can be posted in any of the 5 boxes.
So, the first letter has 5 options, so does the second letter
and so on and so forth for all of the 10 letters.
i.e. 5.5.5….5 (upto 10 times)
= 510.
Here is another way to calculate it:

28. Including "none" as an option, there are 6 choices of appetizer, 11
choices of main meal and 5 choices of dessert. Thus the total number
of choices is 6x11x5=330.
One of these is not a meal though (no appetizer, no main meal and no
dessert), so there are 329 possible meals.
Circular permutations
There are two cases of circular-permutations:-
(a) If clockwise and anti clock-wise orders are different,
then total number of circular-permutations is given by (n-
1)!
(b) If clock-wise and anti-clock-wise orders are taken as not
different, then total number of circular-permutations is
given by (n-1)!/2!
Proof(a):
(a) Let‟s consider that 4 persons A,B,C, and D are sitting
around a round table
Shifting A, B, C, D, one position in anticlock-wise direction, we
get the following agreements:-

29. Thus, we use that if 4 persons are sitting at a round table, then
they can be shifted four times, but these four arrangements will
be the same, because the sequence of A, B, C, D, is same. But
if A, B, C, D, are sitting in a row, and they are shifted, then the
four linear-arrangement will be different.
Hence if we have „4‟ things, then for each circular-arrangement
number of linear-arrangements =4
Similarly, if we have „n‟ things, then for each circular –
agreement, number of linear – arrangement = n.
Let the total circular arrangement = p
Total number of linear–arrangements = n.p
Total number of linear–arrangements
= n. (number of circular-arrangements)
Or Number of circular-arrangements = 1 (number of linear
arrangements)
n = 1( n!)/n
circular permutation = (n-1)!
Proof (b) When clock-wise and anti-clock wise arrangements
are not different, then observation can be made from
both sides, and this will be the same. Here two
permutations will be counted as one. So total
permutations will be half, hence in this case.
Circular–permutations = (n-1)!/2

30. Note: Number of circular-permutations of „n‟ different things
taken „r‟ at a time:-
(a) If clock-wise and anti-clockwise orders are taken as
different, then total number of circular-permutations =
n
Pr /r
(b) If clock-wise and anti-clockwise orders are taken as not
different, then total number of circular – permutation =
n
Pr/2r
Example: How many necklace of 12 beads each can be made
from 18 beads of different colours?
Ans. Here clock-wise and anti-clockwise arrangement s are
same.
18
Hence total number of circular–permutations: P12/2x12
= 18!/(6 x 24)
Restricted – Permutations
(a) Number of permutations of „n‟ things, taken „r‟ at a time,
when a particular thing is to be always included in each
arrangement
= r n-1 Pr-1
(b) Number of permutations of „n‟ things, taken „r‟ at a time,
when a particular thing is fixed: = n-1 Pr-1
(c) Number of permutations of „n‟ things, taken „r‟ at a time,
when a particular thing is never taken: = n-1 Pr.

31. (d) Number of permutations of „n‟ things, taken „r‟ at a time,
when „m‟ specified things always come together = m! x ( n-
m+1) !
(e) Number of permutations of „n‟ things, taken all at a time,
when „m‟ specified things always come together = n ! - [ m!
x (n-m+1)! ]
Example: How many words can be formed with the letters of
the word „OMEGA‟ when:
(i) „O‟ and „A‟ occupying end places.
(ii) „E‟ being always in the middle
(iii) Vowels occupying odd-places
(iv) Vowels being never together.
Ans.
(i) When „O‟ and „A‟ occupying end-places
=> M.E.G. (OA)
Here (OA) are fixed, hence M, E, G can be arranged in 3!
ways
But (O,A) can be arranged themselves is 2! ways.
=> Total number of words = 3! x 2! = 12 ways.

32. ii) When „E‟ is fixed in the middle
=> O.M.(E), G.A.
Hence four-letter O.M.G.A. can be arranged in 4! i.e 24
ways.
(iii) Three vowels (O,E,A,) can be arranged in the odd-
places (1st, 3rd and 5th) = 3! ways.
And two consonants (M,G,) can be arranged in the even-
place (2nd, 4th) = 2 ! ways
=> Total number of ways= 3! x 2! = 12 ways.
(iv) Total number of words = 5! = 120!
If all the vowels come together, then we have: (O.E.A.), M,G
These can be arranged in 3! ways.
But (O,E.A.) can be arranged themselves in 3! ways.
=> Number of ways, when vowels come-together = 3! x
3!
= 36 ways
=> Number of ways, when vowels being never-together
= 120-36 = 84 ways.

33. Number of Combination of „n‟ different things, taken „r‟ at a time
is given by:-
n
Cr= n! / r ! x (n-r)!
Proof: Each combination consists of „r‟ different things, which
can be arranged among themselves in r! ways.
=> For one combination of „r‟ different things, number of
arrangements = r!
For nCr combination number of arrangements: r n
Cr
n
=> Total number of permutations = r! Cr ---------------(1)
But number of permutation of „n‟ different things, taken „r‟ at a
time
= nPr -------(2)
From (1) and (2) :
n
Pr = r! . nCr
or n!/(n-r)! = r! . nCr
n
or Cr = n!/r!x(n-r)!
Note: nCr = nCn-r
n
or Cr = n!/r!x(n-r)! and nCn-r = n!/(n-r)!x(n-(n-r))!

34. = n!/(n-r)!xr!
Restricted – Combinations
(a) Number of combinations of „n‟ different things taken „r‟ at
a time, when „p‟ particular things are always included =
n-p
Cr-p.
(b) Number of combination of „n‟ different things, taken „r‟ at
a time, when „p‟ particular things are always to be
excluded = n-pCr
Example: In how many ways can a cricket-eleven be chosen
out of 15 players? if
(i) A particular player is always chosen,
(ii) A particular is never chosen.
Ans:
(i) A particular player is always chosen, it means that 10
players are selected out of the remaining 14 players.
14
=. Required number of ways = C10 = 14C4
= 14!/4!x19! = 1365
(ii) A particular players is never chosen, it means that 11
players are selected out of 14 players.

35. 14
=> Required number of ways = C11
= 14!/11!x3! = 364
(iii) Number of ways of selecting zero or more things from „n‟
different things is given by:- 2n-1
Proof: Number of ways of selecting one thing, out of n-things
= nC1
Number of selecting two things, out of n-things =nC2
Number of ways of selecting three things, out of n-things =nC3
Number of ways of selecting „n‟ things out of „n‟ things = nCn
=>Total number of ways of selecting one or more things out of
n different things
= nC1 + nC2 + nC3 + ------------- + nCn
= (nC0 + nC1 + -----------------nCn) - nC0
= 2n – 1 [ nC0=1]
1. Examples:
1. Suppose we want to select two out of three boys A, B, C.
Then, possible selections are AB, BC and CA.
Note: AB and BA represent the same selection.
2. All the combinations formed by a, b, c taking ab, bc, ca.
3. The only combination that can be formed of three letters a,
b, c taken all at a time is abc.

36. 4. Various groups of 2 out of four persons A, B, C, D are:
AB, AC, AD, BC, BD, CD.
5. Note that ab ba are two different permutations but they
represent the same combination.
2. Number of Combinations:
The number of all combinations of n things, taken r at a time is:
n n! n(n - 1)(n - 2) ... to r factors
Cr = = .
(r!)(n - r!) r!
Note:
n
i. Cn = 1 and nC0 = 1.
n
ii. Cr = nC(n - r)
Examples:
11 (11 x 10 x 9 x 8)
i. C4 = = 330.
(4 x 3 x 2 x 1)
16 x 15 x 14 16 x 15 x 14
ii. 16C13 = 16C(16 - 13) = 16C3 = = = 560.
3! 3x2x1
1. From a group of 7 men and 6 women, five persons are to be
selected to form a committee so that at least 3 men are there on the
committee. In how many ways can it be done?
A.564 B. 645
C. 735 D.756
2. In how many different ways can the letters of the word
'LEADING' be arranged in such a way that the vowels always
come together?
A.360 B. 480
5040
C. 720 D.

37. 3 how many different ways can the letters of the word
In
.'CORPORATION' be arranged so that the vowels always come
together?
A B
810 1440
. .
C D
2880 50400
. .
4 Out of 7 consonants and 4 vowels, how many words of 3
consonants and 2 vowels can be formed?
A B
210 1050
. .
C D
25200 21400
. .
5. In how many ways can the letters of the word 'LEADER' be
arranged?
A B
72 144
. .
C D
360 720
. .
6 In a group of 6 boys and 4 girls, four children are to be selected. In
how many different ways can they be selected such that at least one
boy should be there?
A B
159 194
. .
C D
205 209
. .
E
None of these
.
7 How many 3-digit numbers can be formed from the digits 2, 3, 5, 6,
7 and 9, which are divisible by 5 and none of the digits is repeated?
A B
5 10
. .
C D
15 20
. .
8 In how many ways a committee, consisting of 5 men and 6 women
can be formed from 8 men and 10 women?
A 266 B 5040

38. . .
C D
11760 86400
. .
E
None of these
.
9 A box contains 2 white balls, 3 black balls and 4 red balls. In how
many ways can 3 balls be drawn from the box, if at least one black
ball is to be included in the draw?
A B
32 48
. .
C D
64 96
. .
E
None of these
.
10 In how many different ways can the letters of the word 'DETAIL'
be arranged in such a way that the vowels occupy only the odd
positions?
A B
32 48
. .
C D
36 60
. .
E
120
.
11 In how many ways can a group of 5 men and 2 women be made
out of a total of 7 men and 3 women?
A B
63 90
. .
C D
126 45
. .
E
135
.
12 How many 4-letter words with or without meaning, can be formed
out of the letters of the word, 'LOGARITHMS', if repetition of letters
is not allowed?
A B
40 400
. .
C 5040 D 2520

39. . .
13 In how many different ways can the letters of the word
'MATHEMATICS' be arranged so that the vowels always come
together?
A B
10080 4989600
. .
C D
120960 None of these
. .
14 In how many different ways can the letters of the word 'OPTICAL'
be arranged so that the vowels always come together?
A B
120 720
. .
C D
4320 2160
. .
E
None of these
.
Answer:1 Option D
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5
men only).
Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
7x6x5 6x5
= x + (7C3 x 6C1) + (7C2)
3x2x1 2x1
7x6x5 7x6
= 525 + x6 +
3x2x1 2x1
= (525 + 210 + 21)
= 756.
Answer:2 Option C
Explanation:

40. The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to
form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
Answer: 3 Option D
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one
letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are
different.
7!
Number of ways arranging these letters = = 2520.
2!
Now, 5 vowels in which O occurs 3 times and the rest are different,
can be arranged
5!
in = 20 ways.
3!
Required number of ways = (2520 x 20) = 50400
Answer:4 Option C
Explanation:

41. Number of ways of selecting (3 consonants out of 7) and (2 vowels
out of 4)
= (7C3 x 4C2)
7x6x5 4x3
= x
3x2x1 2x1
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
= 5!
5 letters among themselves
=5x4x3x2x1
= 120.
Required number of ways = (210 x 120) = 25200.
Answer: 5Option C
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and
1R.
Required number of ways =6!=
Answer:6 Option D
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys
and 1 girl) or (4 boys).
Required number 6
= ( C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
of ways
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
6x5 4x3 6x5x4 6x5
= (6 x 4) x + x4 +
2x1 2x1 3x2x1 2x1
= (24 + 90 + 80 + 15)

42. = 209.
Answer:7 Option D
Explanation:
Since each desired number is divisible by 5, so we must have 5 at the
unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2,
3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4
digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20
Answer:8 Option C
Explanation:
Required number of ways= (8C5 x 10C6)
= (8C3 x 10C4)
8 x 7 x 6 10 x 9 x 8 x 7
= 3 x 2 x 1x 4 x 3 x 2 x 1
= 11760.
Answer9: Option C
Explanation:
We may have(1 black and 2 non-black) or (2 black and 1 non-black)
or (3 black).
Required number of ways= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
6x5 3x2
= 3x + x6 +1
2x1 2x1
= (45 + 18 + 1)
= 64.

43. Answer:10 Option C
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels
and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked
1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 3! = 6.
Answer:11 Option A
Explanation:
7x
7 3 7 3
Required number of ways = ( C5 x C2) = ( C2 x C1) 6 x =
= 2x 3 63.
1
Answer:12 Option C
Explanation:
'LOGARITHMS' contains 10 different letters.
Required number of = Number of arrangements of 10 letters, taking
words 4 at a time.
= 10P4
= (10 x 9 x 8 x 7)
= 5040.

44. Answer:13 Option C
Explanation:
In the word 'MATHEMATICS', we treat the vowels AEAI as one
letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T
occurs twice and the rest are different.
8!
Number of ways of arranging these letters = = 10080.
(2!)(2!)
Now, AEAI has 4 letters in which A occurs 2 times and the rest are
different.
4!
Number of ways of arranging these letters = = 12.
2!
Required number of words = (10080 x 12) = 120960
Answer:14 Option B
Explanation:
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to
form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.

45. ASSIGNMENT
Question 1 The principal wants to arrange 5 students on the
platform such that the boy SALIM occupies the second position
and such that the girl SITA is always adjacent to the girl RITA .
How many such arrangements are possible?
Question 2 When a group- photograph is taken, all the seven
teachers should be in the first row and all the twenty students
should be in the second row. If the two corners of the second row
are reserved for the two tallest students, interchangeable only b/w
them, and if the middle seat of the front row is reserved for the
principal, how many such arrangements are possible?
Question 3 If there are six periods in each working day of a
school, in how many ways can one arrange 5 subjects such that
each subject is allotted at least one period?
Question 4 How many numbers greater than 4,00,000 can be
formed by using the digits 0, 2, 2, 4, 4, 5?
**Question 5 The letters of the word ZENITH are written in all
possible order. How many words are possible if all these words
are written out as in a dictionary? What is the rank of the word
ZENITH?
**Question 6 How many natural numbers not exceeding 4321 can
be formed with the digits 1,2,3 and 4 if the digits can repeat?
**Question 7 Three married couples are to be seated in a row
having six seats in a cinema hall. If spouses are to be seated next
to each other, in how many ways can they be seated? Find also the
number of ways of their seating if all the ladies sit together.
Question8 A boy has three library tickets and 8 books of his
interest in the library. Of these 8, he does not want to borrow
Chem. part II, unless Chem. Par I is also borrowed. In how many
ways can be choose the three books to be borrowed?
Question 9 A polygon has 44 diagonals. Find the number of its
sides.
** Question10 In how many ways can 10 things be equally divided
b/w (i) two persons (ii) two heaps ?
**Question11 In an exam. there are three multiple choice
questions and each question has 4 choices. Find the no. Of ways in
which a student fails to get all answer correct.

46. Question12 In how many ways can the letters of the word
PERMUTATIONS be arranged if the
(i) Words start with P and end with S? (ii) Vowels are all
together ? (iii) T’s are together? (iv) There is no restriction?
(v) There are always 4 letters b/w P & S?
Question 13 Find the no. Of perm. Of n things taken r at a time in
which 2 given things always occur. In how many pers. , they are
always excluded?
**Question 14 There are 10 points in a plane , no three of which
are in the same st. Line excepting 4 points, which are collinear.
Find the (i)
No. Of st. Lines obtained from the pairs of these points , (ii)
no. Of ∆ that can be formed with the vertices as these points.
Question 15 Find the no. of per. Of n different things taken r at a
time such that two specific things occur together.
ANSWERS ( WITH HINTS)
Answer 1 since the boy SALIM occupies the second position, we
have to arrange the remaining 4 students, according to condition
two seats III, IV OR IV , V
May be occupied by SITA & RITA in 4 ways, then
remaining seat may be occupied by 5th student in 1 way only.So
no. Of arrangements = 2.4.1=8.
Answer 2 The remaining 6 teachers can be arranged in the front
row in 6! Ways. (∵ middle seat is reserved for principal). The
remaining 18 students can be arranged in the second row in 18!
Ways. (two corner are reserved for 2 tallest , they can be occupied
2 ways) ∴ total no. Of ways = 6!.(18)!.2!
Answer 3 Five subjects can be allotted 5 periods out of the six
periods in 6P5 ways. Now one period is left and it can be allotted to
any one of the 5 subjects in 5 ways. So, total no. Of ways = 6P5 . 5
= 3600.
Answer 4 When 5 occurs the extreme left position(greater than 4) ,
then no. Of ways = = 30. When 4 occurs at extreme left , then

47. no. Of ways = 5!/2!=60 ( 0,2,2,4,5[2 appears twice]) ∴ total no. Of
ways 90.
Answer 5 (i) total no. Of possible words = 6!=720
(ii) The no. Of words beginning with E = 5! = 120 , similarly for
H ,I, N or T is 120.
The words start with Z will also be 120 and will have their
rank from 601 to 720.Of these 120 words start with Z, the no. Of
words with E in the second place is 120/5 = 24. ZE**** will have
their rank from 601 to 624. Of these 24 words , NO. Of words
with H(or I or N or T) in the third place is 24/2=6
Rank order : ZEH*** , ZEI***, ZEN*** 601 to 606, 607 to
612, 613 to 618 resp., start with ZEN are 613...ZENHIT, 614...
ZENHTI, 615... ZENIHT,
616... ZENITH.
Answer 6 1- digit nos. = 4, 2-digit nos. = 4x16, 3-digit nos. = 64,
4-digit nos. 256. Out of 256 , let us discard those which are greater
than 4321.
(i) 4 at thousand’s place & hundred’s place= 1.1.4.4=16 ,
4 4 1 1
2 2
3 3
4 4
Similarly (ii) 4 at thousand’s place ,3 at hundred’s place, 3 or 4 in
ten’s = 1.1.2.4=8,
(iii) 4 at thousand’s place ,3 at hundred’s place, 2 in ten’s & 2,3,4
at unit’s= 1.1.1.3=3. Total 4-digit no.greater than 4321 are = 27,
answer is 256-27=313.
Answer 7(i) Let A, B, C be three married couples can be seated =
3! & can sit in 2 ways =2! ∴ req. Ways of seating = 3!.2!.2!.2!=48 .
(ii) three ladies can be seated in 4 ways which are at seat
number (1,2,3), (2,3,4), (3,4,5), (4,5,6). They can interchange their
seats =3! Ways. Men can be seated at three remaining seats in 3! ,
so req. No. Of ways = 4.3!.3!= 144 .
Answer 8 case-1 he borrows chem.. part II, chem.. part I and one
more book out of the remaining 6 (8-2) books.

48. Case-2 he does not borrow chem.. part-II and , so
borrow all 3 books out of 7 ∴ total no. Of ways = 6C1 + 7C3=41.
Answer 9 No. Of diagonals = nC2 - n = 44 ( nC2 = no.of st. Lines
of polygon n sides).
Answer 10 (i) 10 things be equally divided b/w two persons,
groups are distinct , 10!/ (5!)2 = 252 (ii) 10 things be equally
divided b/w two heaps no distinction can be made = ½ (252)=126.
Answer 11 Req. No. Of ways= 4.4.4 – 1=63.
Answer 13 (i) 10!/2! (ii) consider 5 vowels as one letter (8!/2!)
.(5!) (iii) consider 2 T’s as one letter = (11)! (iv) 12!/2!
(v) 12 letters in 12 places as P****S****** leaving 4 places in
between , Thus P & S may be filled up in 7 ways, same for S & P ,
remaining 10 letters are arranged in 10!/2! Ways ∴ req. No. of
ways 10!/(3!.2!.2!) = 151200.
Answer 14 No. Of per. Of n things taken r at a time is same as the
no. Of ways of filling up r places with n things, we arrange 2
things which can be arranged in r places in rP2 ways. Also
remaining (r-2) places can be filled up with remaining (n-2) things
in n-2Pr-2 ways.
No. Of per. (2 are included) = rp2 . n-2Pr-2 , No. Of per. (2 are not
included) = n-2Pr
Answer 15 (i) No. Of st. Lines formed by joining 10 points ,
taking 2 at a time = 10C2 = 45. No. Of st. Lines joining 4 points =
4
C2 = 6 , but 4 are collinear , when join pairwise give only one line.
Req. No. of lines = 45 – 6 + 1=40
(ii) No. of ∆s formed by joining the points , taken 3 at a time =
10
C3 = 120, no. of ∆s formed by joining 4 points , taken 3 at a time
= 4C3 = 4
Req. No. of ∆s = 120 – 4 = 116.
Answer 16 Two specific things can be in r places in (r-1) ways and
two things can be arranged among themselves in 2! Ways and the
remaining (n-2) things will be arranged in (r-2) places in n-2Pr-2
ways. Req. Per. = 2! (r-1) . n-2Pr-2 .