StudentAidUnit 7: Permutations and Combinations — Exercise 7 3
11th Class Mathematics · Unit 7: Permutations and Combinations
1.How many arrangements of the letters of the following words, taken all together can be made? (i) PAKISTAN (ii) CURRICULUM (iii) PROBABILITY
Given
(i) Total letters
= 8
A is repeated
2 times
Formula
Number of arrangements
= 8!2! = 403202 = 20160
Given
(ii) Total letters
= 10
C repeated 2, R repeated 2, U repeated 3 times
Formula
Number of arrangements
= 10!2! × 2! × 3! = 362880024 = 151200
Given
(iii) Total letters
= 11
B repeated 2, I repeated 2 times
Formula
Number of arrangements
= 11!2! × 2! = 399168004 = 9979200
(i) 20160, (ii) 151200, (iii) 9979200
2.How many permutations of the letters of the word "BANANA" can be made, if B must be the first letter in each arrangement?
Given
B is fixed at first position
Remaining letters
A, N, A, N, A
Number of remaining letters
= 5
A is repeated 3 times, N is repeated 2 times
Formula
Required number of permutations
= 5!3! × 2! = 1206 × 2 = 12012 = 10
10
3.How many arrangements of the letters of the word TRIGONOMETRY can be made, if each arrangement begins with T and ends with Y?
Given
Total letters in TRIGONOMETRY
= 12
T is fixed at beginning, Y at end
Remaining letters to arrange
= 10
T repeated 1, R repeated 2, O repeated 2 times
Formula
Required number of arrangements
= 10!2! × 2! = 36288004 = 907200
907200
4.Abdullah has a collection of 9 marbles consisting of 4 identical red marbles, 3 identical blue marbles and 2 identical green marbles. If he wants to arrange all of them in a straight row, how many distinct arrangements are possible?
Given
Number of red marbles
= 4
Number of blue marbles
= 3
Number of green marbles
= 2
Total marbles
= 4 + 3 + 2 = 9
Formula
Required number of arrangements
= 9!4! × 3! × 2! = 36288024 × 6 × 2 = 362880288 = 1260
1260
5.In how many different ways can the following persons sit around a round table? (a) 8 persons (b) 7 persons (c) 6 persons
Given
(a) Number of persons
n = 8
By fixing one of the persons
Formula
Required number of ways
= (n-1)! = (8-1)! = 7! = 5040
Given
(b) Number of persons
n = 7
Formula
Required number of ways
= (7-1)! = 6! = 720
Given
(c) Number of persons
n = 6
Formula
Required number of ways
= (6-1)! = 5! = 120
(a) 5040, (b) 720, (c) 120
6.In how many ways can 5 couples sit around a round table if no two women are sitting together?
Given
Here men and women occupy alternate seats
Number of men
= 5
Number of women
= 5
Number of ways of arranging 5 men
= 4! = 24
Number of ways of arranging 5 women
= 5! = 120
Formula
Required number of ways
= 24 × 120 = 2880
2880
7.How many 6-digit numbers can be formed from the digits 7,7,8,8,9,9?
Given
Given digits are
7, 7, 8, 8, 9, 9
Number of digits
= 6
7, 8 and 9 are repeated 2 times
Formula
Required number of arrangements
= 6!2! · 2! · 2! = 7202 · 2 · 2 = 7208 = 90
90
8.15 members of a club form 4 committees of 3, 4, 5 and 3 members so that no member is a member of more than one committee. Find the number of committees.
Given
Total members
= 15
Members of first committee
= 3
Members of second committee
= 5
Members of third committee
= 4
Members of fourth committee
= 3
Formula
Required number of committees
= 15!3! · 5! · 4! · 3! = 1307674368006 × 120 × 24 × 6 = 12612600
12612600
9.The D.C.Os of 11 districts meet to discuss the law-and-order situation in their districts. In how many ways can they be seated at a round table, when two particular D.C.Os insist on sitting together?
Given
Since two Particular D.C.Os insist on sitting together, so consider them as one man for seating.
The total number of men now are
= 10
and they can be seated at a round table in
9! ways
Also, two Particular D.C.Os can occupy their seats in
2! ways
Formula
Therefore, the total number of ways are
9! × 2! = 725760
725760
10.The Governor of the Punjab calls a meeting of 14 officers. In how many ways can they be seated at a round table?
Given
Number of officers
= 14
Formula
Number of ways in which they can be seated at a round table, (fixing one seat), are
= 13!
13!
11.Fatima invites 14 people for a dinner. There are 9 males and 5 females who are seated at two different tables. Guests of one sex sit at one round table and the guests of the other sex sit at the second table. Find the number of ways in which all guests can be seated.
Given
Number of males
= 9
Number of females
= 5
Number of ways that 9 males can be seated at a round table
= 8! = 40320 ways
5 females can be seated at a round table in
4! = 24 ways
Formula
Both males and females can be seated at a round table
= 40320 × 24 = 967680 ways
967680
12.Find the number of ways in which 5 men and 5 women can be seated at a round table in such a way that no two persons of the same sex sit together.
Given
Number of men
= 5
Number of women
= 5
Number of ways the 5 men be seated at round table (fixing one seat)
= 4! = 24 ways
Number of ways the 5 women be seated, each between two men
= 5! = 120 ways
Formula
Required number of ways
= 24 × 120 = 2880 ways
2880
13.In how many ways can 8 keys be arranged in a circular key ring?
Given
Number of keys
n = 8
By fixing one of the keys,
Formula
Required number of ways
= (n-1)!2 = (8-1)!2 = 7!2 = 50402 = 2520
2520
14.How many necklaces can be made from 10 beads of different colours?
Given
Number of beads
= 10
By fixing one of the beads,
Formula
Required number of ways
= (n-1)!2 = (10-1)!2 = 9!2 = 3628802 = 181440
181440