2. Permutation
• The different arrangements
of a given number of things
by taking some or all at a
time, are called
permutations. This is
denoted by nPr.
Combination
• Each of the different
groups or selections
which can be formed by
taking some or all of a
number of objects is
called a combination. This
is denoted by ncr.
3. If we are taking A,B,C and creating two
letter word means…
PERMUTATION
AB,BC,CA,BA,CB,AC
COMBINATION
AB,BC,CA
4.
5. If we are taking A,B,C and creating two
letter word means…
PERMUTATION
AB,BC,CA,BA,CB,AC = 6
nPr = n!/(n - r)!
n=3 r=2
3P2= 3!/(3-2)!
=3*2*1/ 1
= 6
COMBINATION
AB,BC, CA = 3
nCr =n!/(r!)(n - r)!
3C2 =3!/2!*(3-2)!
=3*2*1/2*1*1
=3
8. 5.How many words can be
formed by using all
letters of the word
“BIHAR”?
9. 5.How many words can be formed by using all
letters of the word “BIHAR”?
SOL:
BIHAR = 5 LETTERS
Required number of words = 5P5 (5!)
= 5*4*3*2*1
= 120
10. 6.How many words can
be formed by using all
letters of the word
“LAPTOP”?
11. 6.How many words can be formed by using
all letters of the word “LAPTOP”?
SOL:
LAPTOP = 6 LETTERS
Required number of words = 6P6 /2!
(6!/2!)
Because P is coming twice.so we are putting 2!
in denominator
= 6*5*4*3*2*1/2*1
= 360
12. 7.How many words can
be formed by using all
letters of the word
“BEETROOT”?
13. 7.How many words can be formed by using all
letters of the word “BEETROOT”?
SOL:
BEETROOT = 8 LETTERS
Required number of words = 8P8 /2!*2!*2!
(8!/2!*2!*2!)
Because E,T,O are coming twice. so we are putting
2! in denominator for each
= 6*5*4*3*2*1/2*1*2*1*2*1
= 90
14. VOWELS ALWAYS COME
TOGETHER
8.How many words can be formed
by using all the letters of the
word ”DAUGHTER”, so that the
vowels always come together?
15. 8.How many words can be formed by
using all the letters of the word
”DAUGHTER”, so that the vowels
always come together?
Vowels = A,U,E
Total Letters = 6 (all Vowels are taken as one)
Vowels always come together = (AUE)DGHTR
=(3!)6!
= 3*2*6*5*4*3*2*1
= 4320
16. PRACTICE
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
• C.720
• D.5040
• E.None of these
17. VOWELS NEVER COME TOGETHER
9.How many words can be formed
by using all the letters of the
word “EXTRA” so that the vowels
are never together?
18. 9.How many words can be formed by using all the
letters of the word “EXTRA” so that the vowels
are never together?
Vowels never together = (Possible number of ways
– vowels come together)
Possible number of ways = EXTRA
= 5 LETTERS =5! = 120
Vowels come together = XTR(EA)
= 4! * 2! = 48
Vowels never together = 120 – 48 = 72
19. Combination
10.In how many ways can a cricket 11 be chosen
out of a batch of 15 players?
Total players = 15
Required players = 11
Required number of ways = 15 C 11
20. 11.In how many ways a committee
of 5 members can be selected
from 6men & 5 ladies, consisting
of a 3men & 2ladies?
21. 11.In how many ways a committee of 5 members can
be selected from 6men & 5 ladies, consisting of a
3men & 2ladies?
Sol:
Total men = 6
Required men = 3
Required ways = 6C3
Total ladies = 5
Required ladies = 2
Required ways = 5C2
Total number of ways = 6C3 * 5C2
= (6*5*3/3*2*1)*(5*4/2*1)
= 200
23. 12.There are 12 Yes or No questions.How many
ways can these be answered?
Constant = 2 (YES or NO)
Variable = 12 (12 questions)
Required ways = 212 (Constant variable)
= 4096
24. 13.If 20 people are in a party, then how many
hand shakes will be possible without forming
a cycle?
Sol:
Total people = 20
Number of ways = (n-1)
= 20-1 = 19
25. 14.If 20 people are in a party, then how many
hand shakes will be possible among them?
Sol:
Total people = 20
Number of ways = n(n-1)/2
= 20*19/2
= 190
26. In a birthday party, every person shakes hand
with every other person. If there was a total
of 28 handshakes in the party, how many
persons were present in the party?
• A. 9
• B. 8
• C. 7
• D. 6
East way : Apply options.
27. 15.How many 5 digit numbers
can be formed from the digits
1,2,3,4,5 which are divisible by
4 & repetition allowed?
28. 15.How many 5 digit numbers can be formed from the digits 1,2,3,4,5 which are
divisible by 4 & repetition allowed?
Note: If the last two digits are divisible by 4,then the whole number
is divisible by 4
We have 1,2,3,4,5.By using this we can form 12,24,32,44,52 which
are divisible by 4.(Repetition allowed. so we can put 44 too in
required place).After filling last two digits, we can choose one
number from those 5 numbers and fill remaining places
respectively.
--- --- --- --- --- --- --- --- --- ---
(5C1) (5C1) * (5C1) * (5C1) * (5C1)
(12,24,32,44,52) =5*5*5*5=5
4
29.
30. 16.How many 5 digit numbers
can be formed from the digits
1,2,3,4,5 which are divisible by
4 & repetition not allowed?
31. 16.How many 5 digit numbers can be formed from the digits 1,2,3,4,5 which are
divisible by 4 & repetition not allowed?
Note: If the last two digits are divisible by 4,then the whole number
is divisible by 4
We have 1,2,3,4,5.By using this we can form 12,24,32,44,52 which
are divisible by 4.(Repetition not allowed. so we cant use 44 in
required place).Now we have ony 4 pairs to fill last two
digits.After filling last two digits, we will have ony three
numbers remaining..
--- --- --- --- --- --- --- --- --- ---
(5C1) (5C1) * (5C1) * (5C1) * (5C1)
(12,24,32,44,52) =5*5*5*5=5
4
32.
33. 17.How many number of times will
the digit “7” be written when
listing the integers from 1 to
1000?
34.
35.
36. Note : 2 ,3,4,5,6,7,8,9 – all
will be there 300 times in 1
to 1000
