Inter 2nd Year Maths 2A Permutations and Combinations Solutions Ex 5(c)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(c)

I.

Question 1.
Find the number of ways of arranging 7 persons around a circle.
Solution:
Number of persons, n = 7
∴ The number of ways of arranging 7 persons around a circle = (n – 1)!
= 6!
= 720
Hint: The no. of circular permutations of n dissimilar things taken all at a time is (n – 1)!

Question 2.
Find the number of ways of arranging the chief minister and 10 cabinet ministers at a circular table so that the chief minister always sits in a particular seat.
Solution:
Total number of persons = 11
The chief minister can be occupied a separate seat in one way and the remaining 10 seats can be occupied by the 10 cabinet ministers in (10)! ways.
∴ The number of required arrangements = (10)! × 1
= (10)!
= 36,28,800

Question 3.
Find the number of ways to prepare a chain with 6 different coloured beads.
Solution:
Hint: The number of circular permutations like the garlands of flowers, chains of beads, etc., of n things = $$\frac{1}{2}$$(n – 1)!
The number of ways of preparing a chain with 6 different coloured beads = $$\frac{1}{2}$$(6 – 1)!
= $$\frac{1}{2}$$ × 5!
= $$\frac{1}{2}$$ × 120
= 60

II.

Question 1.
Find the number of ways of arranging 4 boys and 3 girls around a circle so that all the girls sit together.
Solution:
Treat all the 3 girls as one unit. Then we have 4 boys and 1 unit of girls. They can be arranged around a circle in 4! ways. Now, the 3 girls can be arranged among themselves in 3! ways.
∴ The number of required arrangements = 4! × 3!
= 24 × 6
= 144

Question 2.
Find the number of ways of arranging 7 gents and 4 ladies around a circular table if no two ladies wish to sit together.
Solution:
First, arrange the 7 gents around a circular table in 6! ways.
Then we can find 7 gaps between them. The 4 ladies can be arranged in these 7 gaps in 7P4 ways.

∴ The number of required arrangements = 6! × 7P4
= 720 × 7 × 6 × 5 × 4
= 6,04,800

Question 3.
Find the number of ways of arranging 7 guests and a host around a circle if 2 particular guests wish to sit on either side of the host.
Solution:
Number of guests = 7
Treat the two particular guests along with the host as one unit. Then we have 5 guests and one unit of 2 particular guests along with the host.
They can be arranged around a circle in 5! ways.
The two particular guests can be arranged on either side of the host in 2! ways.
∴ The number of required arrangements = 5! × 2!
= 120 × 2
= 240

Question 4.
Find the number of ways of preparing a garland with 3 yellow, 4 white, and 2 red roses of different sizes such that the two red roses come together.
Solution:
Treat that 2nd rose of different sizes as one unit. Then we have 3 yellow, 4 white, and one unit of red roses.
Then they can be arranged in garland form in $$\frac{1}{2}$$ (8 – 1)! = $$\frac{1}{2}$$ (7!) ways.
Now 2 red roses in one unit can be arranged among themselves in 2! ways.
∴ The number of ways of preparing a garland = $$\frac{1}{2}$$ (7!) × (2!)
= $$\frac{1}{2}$$ × 5040 × 2
= 5040

III.

Question 1.
Find the number of ways of arranging 6 boys and 6 girls around a circular table so that
(i) all the girls sit together
(ii) no two girls sit together
(iii) boys and girls sit alternately
Solution:
(i) Treat all 5 girls as one unit. Then we have 6 boys and 1 unit of girls. They can be arranged around a circular table in 6! ways.
Now, the 6 girls can be arranged among themselves in 6! ways.
∴ The number of required arrangements = 6! × 6!
= 720 × 720
= 5,18,400

(ii) First arrange the 6 boys around a circular table in 5! ways. Then we can find 6 gaps between them.
The 6 girls can be arranged in these 6 gaps in 6! ways.

∴ The number of required arrangements = 5! × 6!
= 120 × 720
= 86,400

(iii) Here the number of girls and number of boys are the same.

Hence the arrangements of boys and girls sit alternatively in the same as the arrangements of no two girls sitting together or arrangements of no two boys sitting together.
First, arrange the 6 girls around a circular table in 5! ways. Then we can find 6 gaps between them.
The 6 boys can be arranged in these 6 gaps in 6! ways.
∴ The number of required arrangements = 5! × 6!
= 120 × 720
= 86,400

Question 2.
Find the number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland. In how many of them
(i) all the yellow roses are together
(ii) no two yellow roses are together
Solution:
Hint: The number of circular permutations like the garlands of flowers, chains of beads, etc., of n things = $$\frac{1}{2}$$(n – 1)!
Total number of roses = 6 + 3 = 9
∴ The number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland = $$\frac{1}{2}$$(9 – 1)!
= $$\frac{1}{2}$$ × 8!
= $$\frac{1}{2}$$ × 40,320
= 20,160
(i) Treat all the 3 yellow roses as one unit. Then we have 6 red roses and one unit of yellow roses. They can be arranged in garland form in (7 – 1)! = 6! ways.
Now, the 3 yellow roses can be arranged among themselves in 3! ways.
But in the case of garlands, clockwise arrangements look alike.
∴ The number of required arrangements = $$\frac{1}{2}$$ × 6! × 3!
= $$\frac{1}{2}$$ × 720 × 6
= 2160

(ii) First arrange the 6 red roses in garland form in 5! ways. Then we can find 6 gaps between them.
The 3 yellow roses can be arranged in these 6 gaps in 6P3 ways.
But in the case of garlands, clockwise and anti-clockwise arrangements look alike.
∴ The number of required arrangements = $$\frac{1}{2}$$ × 5! × 6P3
= $$\frac{1}{2}$$ × 120 × 6 × 5 × 4
= 7200

Question 3.
A round table conference is attended by 3 Indians, 3 Chinese, 3 Canadians, and 2 Americans. Find a number of ways of arranging them at the round table so that the delegates belonging to the same country sit together.
Solution:
Since the delegates belonging to the same country sit together, first arrange the 4 countries in a round table in 3! ways.
Now, 3 Indians can be arranged among themselves in 3! ways,
3 Chinese can be arranged among themselves in 3! ways,
3 Canadians can be arranged among themselves in 3! ways,
and 2 Americans can be arranged among themselves in 2! ways.
∴ The number of required arrangements = 3! × 3! × 3! × 3! × 2!
= 6 × 6 × 6 × 6 × 2
= 2592

Question 4.
A chain of beads is to be prepared using 6 different red coloured beads and 3 different blue coloured beads. In how many ways can this be done so that no two blue-coloured beads come together?
Solution:
First, arrange the 6 red-coloured beads in the form of a chain of beads in (6 – 1)! = 5! ways.
Then there are 6 gaps between them. The 3 blue coloured beads can be arranged in these 6 gaps in 6P3 ways.
Then the total number of circular permutations = 5! × 6P3
But in the case of a chain of beads, clockwise and anti-clockwise arrangements look alike.
∴ The number of required arrangements = $$\frac{1}{2}$$ × 5! × 6P3
= $$\frac{1}{2}$$ × 120 × 6 × 5 × 4
= 7200

Question 5.
A family consists of a father, a mother, 2 daughters, and 2 sons. In how many different ways can they sit at a round table if the 2 daughters wish to sit on either side of the father?
Solution:
Total number of persons in a family = 6
Treat the 2 daughters along with a father as one unit. Then we have a mother, 2 sons, and one unit of daughters along with the father in a family.
They can be seated around a table in (4 – 1)! = 3! ways.
The 2 daughters can be arranged on either side of the father in 2! ways.
∴ The number of required arrangements = 3! × 2!
= 6 × 2
= 12