The document discusses permutations and combinations of objects. It provides examples of calculating the number of arrangements and selections of objects in different scenarios. For permutations of distinct objects, it gives the formula nPr = n!/(n-r)!. For combinations of objects, it gives the formula nCr = n!/(n-r)!r!. It then works through multiple examples of using these formulas to calculate permutations and combinations in situations like arranging letters, selecting committee members, and choosing balls from boxes.

1. PERMUTATIONS
.How many two digit nos. can be formed from
digits 1, 2 and 3?
Given: 1,2,3
n = 3
r = 2
n! = n(n-1) (n-2) (n-3). … .3.2.1
If n=3 = 3.2.1 = 6
0! = 1

2. PERMUTATIONS
.How many two digit nos. can be formed from
digits 1, 2 and 3?
Given: 1,2,3
n = 3
r = 2
nPr = n!/(n-r)! = 3!/(3-2)! = 3!/1! =3.2.1/1.1
=6 ways

3. A. Arrangement of n distinct objects taken altogether at a time
nPr = n! = n(n-1)(n-2)..3•2•1
Note: 1! = 1
0! = 1
1. How many ways can four students arrange themselves in a row of 4
chairs.?
nPr= n! If n=r 4! = 4.3.2.1 = 24 nPr = n!/(n-r)! = 4!/(4-4)!=4.3.2.1/0!
=24/1 = 24
2. How many ways can a disc jockey plays the 5 most requested songs of
the day?
5! = 120

4. B. Arrangement of n distinct objects taking r at a time
nPr = n! / (n-r)! , n ≥ r
1. How many ways can 10 students be seated if there only
3 seats available?
10P3 = 10! / (10-3)!
= 10! / 7!
= 10•9•8•7! / 7!
= 10•9•8
= 720

5. 2. There are 14 public schools in a certain district. The district
supervisor plans to visit one school a day for the whole week.
How many ways can she arrange her schedule for the visit?
14P5 = 240240

6. C. Arrangement of n objects with identical elements taken
altogether at a time.
P = n! / a!b!...k! nPr =n!/ r1!.r 2!...
1. How many ways can we arrange the letters of the word:
a. EEL Given: r
P = 3! / 2!1! r1 = 2 E’s
= 3 r2 = 1 L’s
b. TOTOO n = 3
P = 5! / 2!3! 5.4.3!/2.1.3! =20/2 =10
= 120 / 12
= 10

7. 2. How many ways can 3 identical white shirts, 2
identical red shirts and a blue shirt be arranged in a
clothesline?
P = 6! / 3!2!1! = 6.5.4.3!/3!.2.1.1 =6.5.2=60 ways
= 60

8. D. Arrangement of n objects with identical elements taking r
at a time.
1. How many ways can we arrange the letters of the word:
a. EEL taking 2 letters at a time
2 1
E L
2 0 2! / 2! = 1
1 1 2! / 1!1! = 2
3

9. b. TOTOO taking 2 letters at a time
2 3
T O
1 1 2! / 1!1! = 2
0 2 2! / 2! = 1
2 0 2! / 2! = 1
4

10. C. TOTOO taking 3 letters at a time
2 3
T O
1 2 3! / 2! = 3
2 1 3! / 2! = 3
0 3 3! / 3! = 1
7

11. D. WOWWOWWEE taking 3 letters at a time
5 2 2
W O E
1 1 1 3! = 6
1 0 2 3! / 2! = 3
1 2 0 3! / 2! = 3
2 0 1 3! / 2! = 3
2 1 0 3! / 2! = 3
0 1 2 3! / 2! = 3
3 0 0 3! / 3! = 1
0 2 1 3! / 2! = 3
25

12. E. Arrangement of n distinct objects taken altogether at a time in a circular
order.
P = (n-1)!
a. How many ways can a mother, a father and their 3 kids be seated in a circular
order?
P = (5-1)!
= 4!
= 24
b. How many ways can a mother, a father and their 3 kids be seated in a circular
order such that the mother and the father must be seated next to each other?
F M K1 K2 K3
P = (4-1)! 2!
2 4 = 3! 2!
= 12

13. COMBINATIONS

14. How many words can be formed from the letters of the
word PARALLEL?
Given:
r1 = 1 p’s
r2 = 2 a’s
r3 = 1 r’s
r4 = 3 l’s
r5 = 1 e’s
n = 8 letters
nPr = n!/ r1!.r2!... =3360 ways

15. COMBINATIONS-
nCr = nPr / r! = 6/2! = 6/2 = 3 ways
= n!/ (n-r)!/r!
= [n! / (n-r)!] • 1 / r!
nCr = n! / (n-r)!r!
1. How many ways can a committee of three members be selected from
a set of 10 students?
10C3 = 10! / (10-3)!3!
= 120

16. 2. How many possible ways can 3 males and 2 femalesbe selected from a set
of 6 males and 5 females?
3. A box contains 7 distinct red balls, 5 distinct white balls, and 3 distinct green
balls. In how many ways can we select 3 balls such that:
a. they are red
b. two are red and one is white;
7

17. 2. How many possible ways can 3 males or 2 females be selected from a set
of 6 males and 5 females?
6C3 = 6! / 3!3! = 20
5C2 = 5! / 3!2! = 10
6C3 + 5C2 = 20 +10 = 30
3. A box contains 7 distinct red balls, 5 distinct white balls, and 3 distinct green
balls. In how many ways can we select 3 balls such that:
a. they are red
b. two are red and one is white;

18. 2. How many possible ways can 3 males and 2 females be selected from a set
of 6 males and 5 females?
6C3 = 6! / 3!3! = 20
5C2 = 5! / 3!2! = 10
6C3 + 5C2 = 20 +10 = 30
3. A box contains 7 distinct red balls, 5 distinct white balls, and 3 distinct green
balls. In how many ways can we select 3 balls such that:
a. they are red
7C3 = 35
b. two are red and one is white;
7C2 = 21 5C1 = 5
7C2 (5C1) = 21 (5) = 105

19. c. they are of different color;
(7C1) (5C1) (3C1) = 7 (5) (3) = 105
d. there is at least 1 green ball;
3c1 (7c2) = 3(21) = 63
3C1 (7C1) (5C1) = 3(7)(5) =105
3C1 (5C2) = 3(10) = 30
3C2 (7C1) = 3(7) = 21
3C2 (5C1) = 3(5) = 15
3C3 = 1 = 1
235

20. e. none is green.
7C2 (5C1) = 21(5) = 105
7C1 (5C2) = 7(10) = 70
7C3 = 35 = 35
5C3 = 10 = 10
220
or
12C3 = 220