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$3$ boys and $4$ girls have bought tickets for a row of $7$ seats at a party. In how many ways can they arrange themselves if no one sits beside a person of the same sex?

I found the answer $3! \cdot 4!$. However, the book gives the answer $2 \cdot 3! \cdot 4!$. Which is correct?

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  • $\begingroup$ How do you come up with both answers?? $\endgroup$
    – NewGuy
    Mar 9, 2018 at 13:00
  • $\begingroup$ I thought the answer is 3! * 4! But the book answers 2 * 3! * 4! So I just confused. $\endgroup$
    – user9688
    Mar 9, 2018 at 13:09
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    $\begingroup$ The answer is indeed 3! * 4! as there is only one way we can arrange boys and girls { GBGBGBG } $\endgroup$
    – NewGuy
    Mar 9, 2018 at 13:12
  • $\begingroup$ Neither probability nor abstract algebra $\endgroup$
    – King Tut
    Mar 9, 2018 at 13:13
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    $\begingroup$ Your answer is correct for the reason given by @NewGuy. $\endgroup$
    – N. F. Taussig
    Mar 9, 2018 at 14:52

1 Answer 1

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According to Ross, there are $n!$ possible linear orderings of $n$ items (15). Since we are given that girls and boys alternate, odd-numbered positions must be occupied by girls and even-numbered positions must be occupied by boys. Thus the total number of permutations of all seven people is given by the basic principle of counting, which states that if two experiments are to be performed, where experiment 1 can result in any of $m$ possible outcomes and if, for each outcome of experiment 1, there are $n$ possible outcomes of experiment 2, then together there are $mn$ possible outcomes of the two experiments (Ross 2). If we let $m = 3!$ and we let $n = 4!$, then $mn = 3! \cdot4!$.

Work Cited
Ross, Sheldon M., A first course in probability, Boston, MA: Pearson (ISBN 978-0-321-86681-3). ii, 454 p. (2014). ZBL1307.60003.

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