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6 girls are arranged into 2 rows. Given that each row contain at least 1 girl, find the number of the possible combination(s)

My solution: Using the concept of repeated combinations,

$_{2+4-1} C _{4} $

=$_5C_4$

=5

(The answer key says it is 2160, can someone please explain?)

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  • $\begingroup$ "Given that each row contains at least 1 girl.." Isn't it obvious and unavoidable that every row contains exactly 3 girls here? $\endgroup$
    – drhab
    Feb 20, 2018 at 15:32
  • $\begingroup$ $5$ is way too small. Even if I put $5$ in one row and $1$ in the other, that's $6$ ways to do it. $\endgroup$
    – lulu
    Feb 20, 2018 at 15:34
  • $\begingroup$ @drhab The question might allow a row of 4 and a row of 2! $\endgroup$ Feb 20, 2018 at 15:34
  • $\begingroup$ @drhab can't it be 1 for row 1, and 5 for row 2? I also listed it this way : (1,5)(2,4)(3,3)(4,2)(5,1) $\endgroup$
    – Janjan
    Feb 20, 2018 at 15:35
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    $\begingroup$ Mind you, $2160$ seems far too large. I'd have said the answer was $\sum_{i=1}^5\binom 6i=62$. $\endgroup$
    – lulu
    Feb 20, 2018 at 15:35

1 Answer 1

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They are considering different orders of girls to be different. You can imagine putting the girls in one row in $6!=720$ ways then choosing a point to split the rows in five ways. That would give $3600$ possibilities. They may be thinking that $ABC/DEF$ is the same as $DEF/ABC$ but that only deducts $360$ leaving $3240$. I can't get to $2160$.

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  • $\begingroup$ The $3600$ is correct if there is a "front" row and a "back" row. You can get $2160$ if you don't distinguish between the rows. $\endgroup$ Feb 20, 2018 at 16:14
  • $\begingroup$ @BarryCipra: if there isn't a front and back row it seems to divide the $3600$ by $2$ giving $1800$. I can't find the other $360$ then. $\endgroup$ Feb 20, 2018 at 16:21
  • $\begingroup$ @BarryCipra When I considered the possibility that the rows were not distinguishable, I also got $1800$. Let's say Alice is one of the girls. There are $\binom{5}{k}$ ways to choose which $k$ of the other five girls are in the same row as Alice, $(k + 1)!$ ways to arrange the students in her row, and $[6 - (k + 1)]! = (5 - k)!$ ways to arrange the students in the other row, for a total of $$\sum_{k = 0}^{4} \binom{5}{k}(k + 1)!(5 - k)! = 1800$$ $\endgroup$ Feb 20, 2018 at 16:28
  • $\begingroup$ @N.F.Taussig (and Ross), you are both quite right. What I should have said is, you get $2160$ if you never allow the back row to be shorter than the front row. So once you've got $ABCDEF$ as one of the $720$ line-ups, you split after either $A$, $B$, or $C$ (with $DEF$ at least going to the back row). $\endgroup$ Feb 20, 2018 at 16:36

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