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Question

How many ways can $3$ boys and $3$ girls sit in a row if the boys and the girls are each to sit together?

My Approach

Total number of students =$6$, If we consider each student as one cell then total arrangements is $6!$. Now it is given that girls and boys must sit together. So after taking $G$$B$ or $B$$G$ together we are left with 3 cell .

Total number of ways =$3!*2^3=48$,

$2$ option for each pair i.e either $GB$ or $BG$ and we have $3$ pair but the answer is given $72$

Please help me out.

Thanks!

Edit: The full question from A First Course in Probability by Sheldon Ross reads:

(a) In how many ways can $3$ boys and $3$ girls sit in a row?

(b) In how many ways can $3$ boys and $3$ girls sit in a row if the boys and girls are each to sit together?

(c) In how many ways if only the boys sit together?

(d) In how many ways if no two people of the same sex sit together?

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    $\begingroup$ What do you mean when you say "the boys and the girls are each to sit together"? Does that mean everybody is next to at least one person of the opposite sex, only next to people of the opposite sex, or something different? $\endgroup$
    – user49640
    Commented Jun 14, 2017 at 8:53
  • $\begingroup$ no idea ! the question is given in sheldon ross $\endgroup$
    – laura
    Commented Jun 14, 2017 at 8:55
  • $\begingroup$ If the boys and the girls all sit in two distinct group, there are $3!$ ways for the two groups to arrange the $3$ students (in their group), and furthermore, the order can be {Boy group, Girl group} or {Girl group, Boy group}. Therefore, there are $2 * (3!)^2$ ways to arrange the $6$ students, which equals $72$. $\endgroup$
    – Toby Mak
    Commented Jun 14, 2017 at 8:57
  • $\begingroup$ This solution assumes that the boys and girls sit in two separate groups, with only one boy and girl sitting next to each other. $\endgroup$
    – Toby Mak
    Commented Jun 14, 2017 at 8:59
  • $\begingroup$ @user49640 Curiously, "everybody only sits next to people of the opposite sex" and "each sex sits as a group" both give $72$ (answers b and d in the solution sheet) $\endgroup$
    – Henry
    Commented Jun 14, 2017 at 9:01

3 Answers 3

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To get $72$ interpreting this as the boys sitting together and the girls sit together:

  • $3!=6$ ways of ordering the boys among themselves
  • $3!=6$ ways of ordering the girls among themselves
  • $2!=2$ ways of ordering the two groups

Then $6 \times 6 \times 2 = 72$

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  • $\begingroup$ is my approach wrong? $\endgroup$
    – laura
    Commented Jun 14, 2017 at 9:01
  • $\begingroup$ @laura Your approach looks correct for "three couples, with each couple sitting together" but I do not think that is the question $\endgroup$
    – Henry
    Commented Jun 14, 2017 at 9:02
  • $\begingroup$ seeing the question ,it is hard to analyze! $\endgroup$
    – laura
    Commented Jun 14, 2017 at 9:05
  • $\begingroup$ No, that could be one version of the question given the unclear wording of the question. $\endgroup$
    – Toby Mak
    Commented Jun 14, 2017 at 9:05
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    $\begingroup$ @laura It's not your fault. Question makers should always make sure the questions are formulated clearly and as simply as possible. $\endgroup$
    – Toby Mak
    Commented Jun 14, 2017 at 9:06
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The question assumes that boys and girls alternate. As such, we can have $BGBGBG$ and $GBGBGB$. There are 3 options for the first boy and girl, 2 for the second and 1 for the last, so the number of possible arrangements equals:

$$2 \cdot 3! \cdot 3! = 72$$

Edit: the question is indeed not clear at all. According to the solution file provided by Toby Mak, they meant to ask in how many ways the boys and girls can sit in a row, with all girls sitting together and all boys sitting together. In this case we can have $BBBGGG$ or $GGGBBB$, which again results in $2 \cdot 3! \cdot 3! = 72$ possible arrangements.

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  • $\begingroup$ Yes this seems reasonable to me too, but bad by the book to not teach to be more specific in formulating questions. $\endgroup$ Commented Jun 14, 2017 at 8:59
  • $\begingroup$ The solution to this question can be found here: math.utah.edu/~levin/M5010/HW/hw2sol.pdf $\endgroup$
    – Toby Mak
    Commented Jun 14, 2017 at 9:00
  • $\begingroup$ is my approach wrong? $\endgroup$
    – laura
    Commented Jun 14, 2017 at 9:00
  • $\begingroup$ Yes, because you have mixed up the number of ways to order the groups and number of ways to order the children inside the groups in my version of the question. $\endgroup$
    – Toby Mak
    Commented Jun 14, 2017 at 9:01
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    $\begingroup$ I have edited the question to include all four parts of the question from the book. The interpretation that the boys sit together and the girls sit together is the intended question for part (b). The alternating seats interpretation is correct for part (d). Alas, part (b) is poorly worded, which made it difficult to interpret without reading all four parts of the question. $\endgroup$ Commented Jun 14, 2017 at 9:16
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Another way you can think about the problem with boys and girls alternating is with the n choose k :$\binom{n}{k}$ method.

My idea was to group a boy and a girl into one thing. Then we have three entities we want to sort into three seats.

Then the answer becomes:

$$\text{(ways of grouping group 1)}\times\text{(ways of seating group 1)} $$ $$+ \text{(ways of grouping group 2)}\times\text{(ways of seating group 2)}$$ $$ + \text{(ways of grouping group 3)}\times\text{(ways of seating group 3)}$$

$$=\text{(which gender first)}\times\text{(3 boys choose 1)}\times\text{(3 girls choose 1)}\times\text{(3 possible groups to sit in seat 1)}$$ $$+\text{(which gender first)}\times\text{(2 boys left, choose 1)}\times\text{(2 girls left choose 1)}\times\text{(2 possible groups to sit in seat 2) }$$ $$+\text{(which gender first)}\times\text{(1 boy left, choose 1)}\times\text{(1 girl left, choose 1)}\times\text{(1 possible group left to sit in seat 3)}$$ $$=\binom21 \binom31 3+\binom21 \binom21 \binom21 2 +\binom21 \binom11 \binom11 1=72$$

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