Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(c) to find a better approach to solving the problems.
TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(c)
I.
Question 1.
Find the number of ways of arranging 7 persons around a circle.
Solution:
We know that number of circular permutations of ‘n’ dissimilar things (taken all at atime) is (n – 1)!. The number of ways ofarranging 7 persons, around a circle is (7 – 1)! = 6! = 720.
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:
The chief minister always sits in a particular seat, hence, he is arranged in only 1 way.
Now the 10 cabinet ministers in 10 places are arranged in 10! ways.
∴ Total number of ways 1 × 10! = 10!
Question 3.
Find the number of ways of preparing a chain with 6 different coloured beads.
Solution:
We know that the number of circular permutations of hanging type that can be formed using n things is \(\frac{(n-1) !}{2}\).
Hence the number of different ways of preparing the chains with 6 different coloured beads = \(\frac{(6-1) !}{2}=\frac{5 !}{2}\) = 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 3 girls as 1 unit.
This unit along with 4 boys becomes 5 entities.
Number of circular permutations of these units = (5 – 1) ! = 4! = 24
Three girls can be arranged themselves in 3! ways.
∴ Total number of ways = 24 × 3! = 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:
As no two ladies wish to sit together, first we arrange 7 gents.
These 7 gents around a circular table can be arranged in (7 – 1) ! ways i.e., 6! ways.
Now the number of gaps formed are 7.
Number of ways of arranging 4 ladies in these 7 gaps = \({ }^7 \mathrm{P}_4\).
∴ Total number of ways of arranging 7 gents and 4 ladies around a circular table if no two ladies wish to sit together = 6! × \({ }^7 \mathrm{P}_4\).
Question 3.
Find the number of ways of arranging 7 guests and a host around a circle if 2 par-ticular guests wish to sit on either side of the host.
Solution:
Number of guests are 7.
Treat 2 particular guests and host as single unit.
This unit with remaining 5 guests becomes 6 entities.
∴ Number of ways of arranging 6 entities around a circle = (6 – 1) ! = 5!
The 2 particular guest can arrange on either side of the host in 2! ways.
∴ Number of ways of arranging = 5! × 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 2 red roses of different sizes as single unit, which can be arranged in 2! ways. This unit with 3 yellow and 4 white roses of different sizes becomes 8 entities.
The number of ways of preparing a garland with 8 entities are (8 – 1)! = 7! ways.
∴ Number of circular permutations are 7! × 2!
But this being the case of garland, clockwise and anti clockwise arrangements look alike. Hence the required number of ways = \(\frac{1}{24}\) × 7! × 2! = 5040.
III.
Question 1.
Find the number of ways of arranging 6 boys and 6 girls around a circular fable so that
i) all the girls sit together
ii) no two girls sit together
iii) boys and girls sit alternately.
Solution:
Given 6 boys and 6 girls.
i) All the girls sit together :
Treat all the girls as 1 unit. Then we have 6 boys and 1 unit of girls.
They can be arranged around a circular table in 6! ways.
Again, the 6 girls can be arranged themselves in 6! ways.
Total number of arrangements = 6! × 6!.
ii) No two girls sit together :
As no two girls sit together, first we arrange 6 boys around a circular table.
This can be done in 5! ways.
Then we can find 6 gaps between them.
6 girls in these 6 gaps can be arranged in 6! ways.
∴ The number of arrangements = 5! × 6!.
iii) Boys and girls sit alternately :
As number of boys is equal to number of girls, the arrangement of boys and girls sit alternately is same as no two girls sit together.
First arrange 6 boys around circular table.
This can be done in 5! ways.
Then we find 6 gaps.
Arranging 6 girls in these 6 gaps can be done in 6! ways.
∴ Total number of arrangements = 5! × 6!.
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:
Given 6 red roses and 3 yellow roses of different sizes.
∴ Total number of roses are 9.
∴ The number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland = \(\frac{(9-1) !}{2}=\frac{8 !}{2}\) = 20160.
i) All the yellow roses are together :
Treat yellow roses as one unit.
Then this unit with 6 red roses can have the circular permutations in (7 – 1)! = 6! ways.
Now 3 yellow roses can be arranged themselves in 3! ways.
But in the case of garlands, clockwise and anti clockwise arrangements look alike.
∴ The number of arrangements = \(\frac{6 ! \times 3 !}{2}\) = 2160.
ii) No two yellow roses are together :
As no two yellow roses are together, first arrange 6 red roses in garland form.
This can be done in 5! ways.
Then we find 6 gaps. Arrangement of 3 yellow roses in these 6 gaps can be done in \({ }^6 P_3\) ways.
But in the case of garlands, clockwise and anti clockwise arrangements look alike.
∴ The number of arrangements = \(\frac{1}{2} \times 5 ! \times{ }^6 P_3\) = 7200.
3 Chinese can be arranged themselves in 3! ways.
3 Canadians can be arranged themselves in 3! ways.
2 Americans can be arranged themselves in 2! ways.
∴ The number of required arrangements are 3! × 3! × 3! × 3! × 2! = 2592.
Question 4.
A chain of beads is to be prepared using 6 different red coloured beads and 3 differ-ent blue coloured beads. In how many ways can this be done so that no two blue coloured beads come together.
Solution:
Given 6 different red coloured beads and 3 different blue coloured beads.
As no two blue coloured beads come together, first arrange 6 red coloured beads in the form of chain.
This can be done in (6 – 1)! = 5! ways.
Then 6 gaps are formed between them.
Now arrangement of 3 different blue coloured beads in these 6 gaps can be done in hP3 ways.
Then total number of circular permutations are \({ }^6 \mathrm{P}_3\) × 5!.
But, this being the case of chain, clockwise and anti clockwise look alike.
Hence required number of ways = \(\frac{1}{2} \times{ }^6 \mathrm{P}_3 \times 5\) = 7200.
Question 5.
A family consists of father, mother, 2 daugh¬ters 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:
Treat 2 daughters and father as 1 unit.
This unit with mother and 2 sons becomes 4 entities.
Number of ways can 4 entities arranged around circular table are (4 – 1) ! = 3! ways.
Two daughters on either side of the father can be arranged in 2! ways.
∴ Required number of arrangements = 2! × 3! = 12.