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Here is the problem:

A group of $20$ students, including $3$ particular girls and $4$ particular boys, are to be lined up in two rows with $10$ students each. In how many ways can this be done if the $3$ particular girls must be in the front row while the $4$ boys must be in the back row?

Below is my solution, which is wrong according to the answer in the book. The question is where is the mistake?

First, choose $7$ students to be with the girls in the first row. There are $\binom{13}{7}$ ways to do this. The remaining six students are with the boys. There are $3!$ ways to arrange the girls and $4!$ ways to arrange the boys. If you treat the groups of boys and girls as single objects, you have $8$ objects in the first row and $7$ objects in the second row. Then the total number of ways to arrange the students as required is $$\binom{13}{7} \cdot 3! \cdot 7! \cdot 4! \cdot 8!$$

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  • $\begingroup$ Your reasoning looks correct. Have you compared the numerical values of your answer and the book's answer? $\endgroup$ – N. F. Taussig Jun 4 '15 at 11:47
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    $\begingroup$ You have assumed that the girls must be together, so, too, the boys, but the question only sprcifies their rows $\endgroup$ – true blue anil Jun 4 '15 at 11:57
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It seems to me that you've accounted for all of the ways the particulars can be ordered within the set of particulars, but not for how they might be interspersed with the row at large. My answer would have been ${13 \choose 7} \times 10! \times 10!$ -- that is, I worry about the ordering of the rows at the end rather than ordering the particulars and non-particulars separately. Is that the answer they were looking for?

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  • $\begingroup$ Sorry for the late response. Went to sleep shortly after posting this. $\endgroup$ – alexgiorev Jun 4 '15 at 14:01
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${13\choose 7}$ choices of people to go with the 3 particular girls for the front row.

The back row gets chosen automatically.

Permute each row separately to get result: ${13\choose 7}$ $\cdot$ $(10!)^2$

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  • $\begingroup$ Be careful. Not all $10!$ arrangements of the front row keep the girls together and not all $10!$ arrangements of the back row keep the boys together. $\endgroup$ – N. F. Taussig Jun 4 '15 at 10:04
  • $\begingroup$ @Taussig: Where does it say that the girls have to be together ? Quoting the text, "the 3 particular girls must be in the front row while the 4 boys must be in the back row" $\endgroup$ – true blue anil Jun 4 '15 at 11:51
  • $\begingroup$ Good point. I read too much into the question. You are correct. $\endgroup$ – N. F. Taussig Jun 4 '15 at 12:05
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    $\begingroup$ I don't see how this is different from my answer 7 hours earlier. $\endgroup$ – Shane Jun 4 '15 at 13:27
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    $\begingroup$ @Shane: Essentially the same, I gave another wording because OP went to sleep ! Upvoted $\endgroup$ – true blue anil Jun 4 '15 at 15:14
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I believe your first mistake is that you've said there are $\binom{13}{7}$ ways to arrange the other seven students given that you've taken the group of the three (rather fussy!) girls into account. For convenience, let's name the 20 students A,B,C,...,R,S,T (so girls=$\{A,B,C\}$ and boys=$\{Q,R,S,T\}$). $\binom{13}{7}$ gives the number of ways the remaining 13 people can be arranged seven at a time when order does not matter. But if you lined up a particular group of 7, say D,E,F,G,H,I,J then this is the same as E,D,F,G,H,I,J but these are two distinct ways of arranging the students. My approach would be consider the three groups on their own. As you said there are 6 ways to arrange the girls and 24 to arrange the boys. For the remaining 13 students, where order does matter, we have $^{13}P_7=1716$ ways of arranging them. So the answer is $1716\times 24\times 6 = 247104$ ways which we then multiply by two since there are two lines. Hope this helps.

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    $\begingroup$ You have not taken into account the permutations within each row. See the other answers. $\endgroup$ – N. F. Taussig Jun 4 '15 at 12:08

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