Probablity - Standing in Line

[garbage100]

Problem1:
5 boys and 5 girls are made to stand in a line such that the boys and girls alternate. In how many ways can they be made to stand in the line

Problem2:
5 boys and 5 girls are made to stand in a line such that no two girls stand next to each other. In how many ways can the line be formed?

Are these two questions different? If so how?

Official Answer to follow.

[sdasar]

I don't see any difference between them.

The answer is 5! * 5 !.

[jmeasp]

yes i suppose they are diff:
1.As sdasar has given above
2.The q stem says that no two girls can be together but boys can be, so besides above ways of 5!*5!,we will also have additional six ways

-b-b-b-b-b-,these six spaces can be filled by 5 girls in 6 ways.
does it make sense?

[sdasar]

I see your point. Even then I would say both questions are same, because in the first question it doesn't say who starts the line (boy or girl). As long as they alternate, that is good enough.

So I need to adjust my answer. But I am not sure that all we get is additional 6 ways. In fact I think the answer should be 5! * 5! * 2, because for every interlacing combination of b-b-b-b-b and g-g-g-g-g you have another possibility by deciding to start the line with opposite member. That is for very bgbgbgbgbg , we have gbgbgbgbgb combination that can be deemed different.

[800Bob]

The two questions are not the same. The first question admits only bgbgbgbgbg and gbgbgbgbgb. The second question also admits gbbgbgbgbg, gbgbbgbgbg, gbgbgbbgbg, and gbgbgbgbbg.

So the answer to question 1 is 2*5!*5!, and the answer to question 2 is 6*5!*5!.

[sdasar]

Thanks 800bob. I agree. I didn't see the two boys together scenario.

jmeasp, I misunderstood when you said six additional ways. Now I know you meant mutiplying by 6. Thanks.

[garbage100]

Thanks.