How many ways can 6 boys and 4 girls stand in a row if the girls must not be together?

  1. Arrange all 6 boys in a row . --> ways to do this is 6!

  2. This leaves 7 gap in between the boys to place 4 girls in. -> $_7C_4$

  3. The girls in between the boys can have 4! ways to arrange them.

therefore, answer is $6! \cdot _7C_4 \cdot 4! $

why am i wrong ?

the right answer is $6! \cdot _7C_4 \cdot 4! + 6! \cdot _7C_2 \cdot 4! + 6! \cdot _7C_2 \cdot 4!$

  • $\begingroup$ It would help if you supplied the correct answer. $\endgroup$ – herb steinberg Nov 24 '18 at 18:31
  • $\begingroup$ Maybe the boys (and girls) are not supposed to be distinguishable from eah other? What were you told the right answer is? $\endgroup$ – Bram28 Nov 24 '18 at 18:34
  • $\begingroup$ Your answer sounds correct to me, I do not know why you think it might be incorrect. Reiterating what herbsteinberg said before, please include why you think it is incorrect and what the "correct" answer is. It could simply be that the answers are the same but written in a different format. $\endgroup$ – JMoravitz Nov 24 '18 at 18:35
  • $\begingroup$ It looks like "girls must not be together" is supposed to mean that you can't have all four girls together rather than the stronger restriction of no two girls together. Poorly worded question. $\endgroup$ – Ned Nov 24 '18 at 20:17
  • $\begingroup$ @Ned so how does that explain the other 2 sets which are added to the working i have ? $\endgroup$ – Erikien Nov 25 '18 at 12:56

Say we have girls $A,B,C,D$. The numbers that $X,Y$ are together is $2\cdot 9!$, then $X,Y$ and $Y,Z$ are together is $2\cdot 8!$. Now we use PIE.

Then at least two are together is $${4\choose 2}\cdot 2\cdot 9!-24\cdot 8! = 84\cdot 8!$$

So the answer is $$10!-84\cdot 8! = 8! (90-84) = 6\cdot 8!$$


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