Permutations and combinations

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