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?

  • $\begingroup$ You can't choose any $6$...the unfilled gap must be either first or last. $\endgroup$
    – lulu
    Mar 18, 2017 at 11:34
  • $\begingroup$ 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. $\endgroup$ Mar 18, 2017 at 11:34
  • $\begingroup$ Oh right! Thank you so much. $\endgroup$
    – Arishta
    Mar 18, 2017 at 11:35

2 Answers 2


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)


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