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$6$ boys and $6$ girls form a line with boys and girls alternating.Find the number of ways of making the line.

Answer $2(6!)^2$

I was trying to solve this question using Permutation and Combination

$$\square B \square B \square B \square B \square B \square B \square$$

Blank boxes shown above are for girls.Therefore girls can be chosen in $C(7,6)$. Now boys and girls can be arranged in $6!$ ways.

Therefore, it's answer must be $C(7,6) \cdot 6! \cdot 6!$.

Please point out my mistake.

Note- I got correct answer using Permutation and logical thinking but I want to solve by Combination too.

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  • $\begingroup$ Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 12 '18 at 15:04
  • $\begingroup$ What you calculated is the number of ways of seating six boys and six girls in a row in such a way that no two of the girls sit in adjacent seats. $\endgroup$ – N. F. Taussig Sep 12 '18 at 15:06
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Using your approach, you are selecting 6 of the 7 boxes, where as you want to select either the first 6 or the last 6 but no other selection is valid. For example, the $\binom{7}{6}$ selections includes one where there is no girl in the middle and so there are two boys together which is not permitted by the question.

There are two broad possibilities, the line starts with a girl or a boy. Hence the $2$.

Once you pick the starting gender, there are 6 positions for the girls and 6 for the boys. There are $6!$ ways of placing the girls and similarly for the boys.

Hence the total is $2 (6!) (6!)$.

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    $\begingroup$ thanks sir for giving your precious time $\endgroup$ – Bobbey Ashley Sep 12 '18 at 14:59
  • $\begingroup$ @BobbeyAshley: Counting is surprisingly tricky. It is good to look at other ways of counting, this is how we learn. $\endgroup$ – copper.hat Sep 12 '18 at 15:03
  • $\begingroup$ Yes,Sir really I was just bluffed by this. $\endgroup$ – Bobbey Ashley Sep 12 '18 at 15:08
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Therefore girls can be chosen in C(7,6)

Note that C(7,6) (out of seven, pick six to include) is the same as C(7,1) (out of seven, pick one to exclude), which is just 7. There are seven boxes, so it doesn't make sense to say that this is how many ways there are of choosing girls. It would make more sense to say that this is how many ways of choosing boxes for girls. But that still isn't correct, because every box between two boys must be chosen (otherwise, you would end up with two boys next to each other). So there are only two choices for which box to exclude: either the first box can be excluded, or the last.

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  • $\begingroup$ Thanks sir for pointing out my silly and stupid mistake $\endgroup$ – Bobbey Ashley Sep 12 '18 at 14:57
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By using combination, you cannot use $C(7,6)$ because this means the girls can choose any 6 from among those 7 blank seats. What if 2 of the chosen boxes were the two ends? So one of the middle boxes must not be occupied by a girl, and hence, two boys become adjacent (which you do not want to happen).

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  • $\begingroup$ seats = boxes I mean.. $\endgroup$ – Admuth Sep 12 '18 at 14:50
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For boys and girls each disposition has $6!$ possibilities. You are forced to put one girl and then one boy, so you obtain that answer given the fact that you choose if you start with a boy or a girl. You cannot use combinations simply because in this case the ordering is significative. Also Idk where you get that $7$ from

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  • $\begingroup$ "Also Idk where you get that 7 from" There are seven places to put six girls (a girl can be before the first boy, between the first boy and the second boy, etc.). So we're choosing six out of the seven places to put girls. The error comes in thinking that all combinations of placement are valid. $\endgroup$ – Acccumulation Sep 12 '18 at 14:54
  • $\begingroup$ Oh, I see. I did that arranging both separately and then combining them together one at a time. $\endgroup$ – tommy1996q Sep 12 '18 at 15:02

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