# Combinations with restrictions.

• Jul 19th 2011, 09:39 PM
downthesun01
Combinations with restrictions.
Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated
(a) with no restrictions?
8!=40320

(b) if each couple is to sit together?
2(4!)=48

(c) if all the men sit together to the right of all the women?
4!+4!=48

I just want to make sure that they're right. Thanks
• Jul 20th 2011, 04:39 AM
mr fantastic
Re: Quick combination question
Quote:

Originally Posted by downthesun01
Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated
(a) with no restrictions?
8!=40320

(b) if each couple is to sit together?
2(4!)=48

(c) if all the men sit together to the right of all the women?
4!+4!=48

I just want to make sure that they're right. Thanks

I think the + should be a x in your answer to part (c), and therefore the answer should be .....
• Jul 20th 2011, 04:46 AM
Plato
Re: Combinations with restrictions.
Quote:

Originally Posted by downthesun01
Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated
(a) with no restrictions?
8!=40320

(b) if each couple is to sit together?
2(4!)=48

(c) if all the men sit together to the right of all the women?
4!+4!=48

Part (b) is a gross under-count. The answer should be $2^4(4!)$. Now you tell us why that is correct.

As Mr F points out, part (c) is multiplicative and is missing one factor.
• Jul 20th 2011, 09:49 PM
downthesun01
Re: Combinations with restrictions.
I see why b and c are wrong now. For B, I thought it was enough to multiply by 2, but after thinking about it, it makes sense that I have to multiply by 2 four times (once of each couple).

C was more of a slight mental error on my part.

Thank you for the help