# Number of ways in which boys and girls sit alternately if six boys and six girls sit randomly?

Six boys and six girls sit in a row randomly. What is the total number of ways in which the boys and the girls sit alternately?

My attempt: Consider these six seats _ _ _ _ _ _

The number of ways to arrange 6 boys in 6 places is 6! Now 7 gaps are created between these 6 seats. So, we can select any 6 of these 7 gaps and make girls sit there. There are 7C6 * 6! ways to do that (since the girls can shuffle amongst themselves).

Hence the total number of ways to make boys and girls sit alternately should be 6! 7C1 * 6! but the answer is 2*6!*6!.

What am I missing here?

• You can't choose any $6$...the unfilled gap must be either first or last.
– lulu
Commented Mar 18, 2017 at 11:34
• Note "alternately". Only two of the seven ways to pick six of the gaps result in an alternating pattern, the rest have two boys next to each other. Commented Mar 18, 2017 at 11:34
• Oh right! Thank you so much. Commented Mar 18, 2017 at 11:35

You are implicitly assuming that there are 13 seats for $6+6=12$ people. It is likely that the original question is making the implicit assumption that there are only 12 seats. In this latter case, when a boy sits first, there $6! \cdot 6!$ ways to sit the others. Multiply this by 2 to cover the case where the first seater is a girl.

You are over-counting.

You are also counting the cases where extreme 2 gaps are both occupied and out of 5 gaps in middle 4 are occupied.

For example, let's first arrange the boys: B_B_B_B_B_B = 6! ways.

We are left with 7 gaps so we select 6 among them and permute the girls so (7C6 6! ways.)

Hence according to you answer is 6! . 7C6 . 6! = 7.6!.6!

But it also counts the cases like GBGBGBBGBGBG two boys are sitting together and alternate pattern is disturbed so we eliminate these unfavorable cases by bijection. So, Unfavorable ways= 6! . 2C2 . 5C4. 6! = 5.6!.6! So hence total ways of favourable permutations = 7.6!.6! - 5.6!.6! =2.6!.6! (The required answer)