The problem is, there are 8 boys and 2 girls and the girls are not allowed to sit on either end. In my mark scheme it states that the number of possible combinations possible is: $10!-2*8!-2*2*7*8!$. Personally speaking I did $8*7*8!$ which yield the same answer but I do not get the mark-scheme method. Also I think it makes much more sense to go:$10!-2*8!-2*2*9!$ where $2*9!$ is the possible ways a girl can sit on one end , and $2*2*9!$ is the possibility of one girl sitting on one end at both ends.

Can anyone help me understand where the $2*2*7*8!$ and $2*8!$ comes from?

  • $\begingroup$ Not the same answer: $10!-2*8!-2*2*7*8!=2419200\neq2257920=8*7*8!$. $\endgroup$ – drhab Jan 9 at 13:23
  • $\begingroup$ The mark scheme should say $10!-2*8!-2*2*8*8!$. $\endgroup$ – Jaap Scherphuis Jan 9 at 14:02
  • $\begingroup$ I think the mark scheme had a typo which tripped me up, thank you for the help. $\endgroup$ – Aurora Borealis Jan 10 at 1:59

First of all, I like your approach, which is pretty straightforward. For two girls we have $8$ different spots, and then $8$ boys can be spotted randomly. Hence, there are exactly $$ P(8,2) \times 8!$$ ways.

Now think differently.

step 1. Compute all the permutations of $10$ people, which gives us $10!$

step 2: compute all the permutations where two girls occupy two end spots. In this case, we have $2! \times 8!$ permutations.

step 3. compute the number of permutations where exactly one girl takes an end spot. In this case, you fix exactly one girl at one of the end spots, and another girl sits somewhere at a middle spot. Now we have $$4\times 8 \times 8!$$ permutations.

Note that there are no common permutations between step 2 and step 3. Hence, subtracting the total number of permutations of step 2 and step 3 from step 1's permutations should yield the result. Varify that both solutions give the same answer.

One comment about your second approach: when you trying to solve the problem using your second method whenever you're fixing a girl at some end spot, please make sure that another girl does not occupy another end spot. This is just to make sure there is no overcounting with step 2. By the way, you definitely can overcount and balance things by inclusion-exclusion, but I would not recommend doing so for this kind of straightforward problem.

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  • $\begingroup$ His method can also be explained by choosing the two boys to put on the ends, and then filling the middle 8 seats in any order. $\endgroup$ – Jaap Scherphuis Jan 9 at 13:48
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    $\begingroup$ Thanks for the answers i now get it completely $\endgroup$ – Aurora Borealis Jan 10 at 2:00

You calculation of $8*7*8!$ is a perfectly valid answer too.

Their calculation is $10!-2*8!-2*2*8*8!$. (Note you have a typo.)
Here $2*8!$ counts all the ways of having the two girls on the ends and the 8 boys in any order in the middle.
The $2*2*8*8!$ is all the ways of having exactly one girl at one end, constisting of $2$ choices of end where the girl is, $2$ choice of which girl to place there, $8$ choices of boy to place at the other end, and $8!$ ways to fill the middle $8$ seats (with the remaining 7 boys and 1 girl).

Your proposed term $2*9!$ counts the number of ways you can have a girl on the left end and fill the other 9 seats in any order, but this includes the possibility of having the other girl on the right end. So this is all cases with the girls at both ends and all cases with a girl only on the left end. To avoid double counting, the next term should count only those cases with a girl on the right but not on the left. That would be $7*2*8!$ (7 choices for the boy on the left, 2 choices for the girl on the right, and then arrange the middle 8).
So your answer could be $10!-2*9!-7*2*8!$, and this gives the same result as other two methods.

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  • $\begingroup$ Thank you for the feedback helped me out $\endgroup$ – Aurora Borealis Jan 10 at 2:00

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