3!*3!*2! = 6*6*2 = 72

Answer C

Bunch up the boys together and the girls together. Now we have to arrange 2 groups which we can do in 2! ways.

The boys within themselves can be arranged in 3! ways and the girls within themselves can be arranged in 3! ways.

Total arrangements = 2 * 3! * 3! = 72

Answer (C)

For more on arrangements with constraints, check:
https://www.veritasprep.com/blog/2011/1 ... ts-part-i/
https://www.veritasprep.com/blog/2011/1 ... s-part-ii/
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Karishma
Veritas Prep GMAT Instructor

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We need to determine how many ways to arrange 3 boys and 3 girls in a row of six chairs when the boys must be seated together and the girls must be seated together. Since order matters, we have a permutation problem.

We can arrange the boys and girls as follows:

[B1)-B2)-B(3)] - [G(1)-G(2)-G(3)]

(Note: Since the boys must sit together and the girls must sit together, we have included the boys in one bracket and the girls in one bracket.)

Since we have 2 brackets, we can arrange those 2 brackets in 2! or 2 ways. However, we must account for the individual arrangements of the boys and the girls. The boys can be arranged in 3! or 6 ways and the girls can be arranged in 3! or 6 ways. Thus, the group can be arranged in 2 x 6 x 6 = 72 ways.

Answer: C
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