Question

If four boys are to be seated for a dinner such that no two girls and no two boys sit together.Find the number of ways in which they can be seated?

• A. $144$
• B. $244$
• C. $300$
• D. $225$

Answer 1

The correct option is A $144$

If four boys and three girls are to be seated for a dinner such that no two girls and no two boys sit together. Then the possible number of ways can be :

1st place	2nd place	3rd place	4th place	5th place	6th place	7th place
Boy	Girl	Boy	Girl	Boy	Girl	Boy

So, now for first place we have 4 choices as there are 4 boys, for second place we have 3 choices as there are 3 girls, for third place we have 3 choices as there are 3 boys remaining and so on.

So, total number of possible ways $=4\times 3\times 3\times 2\times 2\times 1\times 1=4!\times 3!=24\times 6=144ways$