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this is a question that I am struggling with. I thought I did it correctly, but the answer key used a different method.

The question: "How many ways can 5 boys and 5 girls stand in a line if the girls and boys alternate throughout the line?"

The way I worked through the problem:

Since there were ten people in total and they were standing in random order, I just said the answer was 10!.

The way the answer key worked through the problem:

5! * 5! * 2 = 28,800

Am I missing something or did I just interpret the problem wrong? (thanks)

Nicole K
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1 Answers1

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The way you're doing it, you allow for all of the ten people to be in any order in the line. This neglects the restriction of "alternating boys and girls."

To explain the answer key method, you can envision the proper lines as being thus:

$$BGBGBGBGBG \;\;\;\;\; \text{or} \;\;\;\;\; GBGBGBGBGB$$

where $B$ signifies a boy and $G$ signifies a girl. These are the only formats allowed in this scenario.

However, you can change up the order of the boys anyway you want, so long as they remain in the boys' positions, granting you $5!$ arrangements. Independently, you can also rearrange the girls any way you want, giving $5!$ further arrangements. Finally, this logic holds for both line types allowed, so we double the end result.

Thus, the answer is $5! \cdot 5! \cdot 2$.

PrincessEev
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  • OH!! This makes much more sense. Another problem asks what would happen if these people sad at a circular table. What would I do here? – Nicole K Mar 18 '20 at 18:26
  • Well now arrangement of the $5$ boys around the table would be $4!$ ways and then in the $5$ gaps we can arrange the girls in $5!$ ways. Finally, we divide by two to avoid counting the same arrangement twice in both clockwise and anticlockwise directions – sai-kartik Mar 18 '20 at 18:29
  • The answer key states that the answer would be 5!⋅5! – Nicole K Mar 18 '20 at 18:45