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  • Find the number of ways in which 6 boys and 4 girls can be seated in a row so that (i) Find the number of ways in which 3 boys and 2 girls can be seated in a row so that exactly two girls sit toget

Find the number of ways in which 6 boys and 4 girls can be seated in a row so that (i) Find the number of ways in which 3 boys and 2 girls can be seated in a row so that exactly two girls sit together. ii) Find the number of ways in which 6 boys and 3 girls can be seated in a row so that at least two girls sit together. iii) Find the number of ways in which 5 boys and 4 girls can be seated in a row so that no two boys sit together. 

 

Option: 1

(i) 18 ways (ii) 678,566 ways (iii) 5,670 ways


Option: 2

(i) 14 ways (ii) 568,870 ways (iii) 2,480 ways


Option: 3

(i) 16 ways (ii) 450,520 ways (iii) 5,660 ways


Option: 4

(i) 12 ways (ii) 358,560 ways (iii) 2,880 ways


Answers (1)

best_answer

(i) Find the number of ways in which 3 boys and 2 girls can be seated in a row so that exactly two girls sit together:

To solve this, we can treat the group of two girls sitting together as a single entity. Therefore, we have 3 entities to arrange: the group of two girls, the remaining one girl, and the three boys.

The group of two girls can be arranged in \mathrm{P(2,2)=2 !=2} ways.
The remaining one girl can be arranged in \mathrm{P(1,1)=1} way.
The three boys can be arranged in \mathrm{P(3,3)=3 !=6} ways.
Therefore, the total number of ways is \mathrm{2 \times 1 \times 6=12}.

(ii) Find the number of ways in which 6 boys and 3 girls can be seated in a row so that at least two girls sit together:

To solve this, we can calculate the total number of ways without any restrictions and subtract the number of ways where no girls sit together.

The total number of ways without any restrictions is \mathrm{P(9,9)=9 !=362,880}.
To find the number of ways where no girls sit together, we can arrange the boys and girls separately in an alternating manner.

The boys can be arranged in \mathrm{P(6,6)=6 !=720} ways.
The girls can be arranged in \mathrm{P(3,3)=3 !=6} ways
Therefore, the number of ways where no girls sit together is  \mathrm{720 \times 6=4,320}.

So, the number of ways where at least two girls sit together is 362,880 - 4,320 = 358,560.

(iii) Find the number of ways in which 5 boys and 4 girls can be seated in a row so that no two boys sit together:

To solve this, we can treat the boys and girls as distinct entities and arrange them separately.

The boys can be arranged in \mathrm{P(5,5)=5 !=120} ways.
The girls can be arranged in \mathrm{P(4,4)=4 !=24} ways. Therefore, the total number of ways is \mathrm{10 \times 24= 2,880}.

So, the answers are:

(i) 12 ways.

(ii) 358,560 ways.

(iii) 2,880 ways.

 

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