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Permutations Questin

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by emw13 » Wed Sep 30, 2009 5:59 pm
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?
b. If each couple is to sit together
c. If all the men sit together on the right of all the women.


THANKS>
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Source: — Problem Solving |

by sanjana » Wed Sep 30, 2009 7:17 pm
This is what I have for the 3 Variations

1)This is as good as seating 8 ppl in a row of 8 seats without any restriction.
No of arrangements = 8! = 40320

2)
This could be viewed as

WHWHWHWH
or
HWHWHWHW

Now for WHWHWHWH
Number of ways to select 1 W - 4
For this selected wife there is only 1 husband
Therefore,the total number of arrangements is
(4x1)X(3X1)X(2X1)X(1X1) = 24
Multiply this by 2 as we could have HWHWHWHW also
Therefore the answer required is 48

3)Number of ways of WWWWHHHH
4!X4! = 24X24 = 576

Please can you confirm the answers?
Thanks!
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Re: Permutations Questin

by tohellandback » Wed Sep 30, 2009 9:18 pm
emw13 wrote: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?
b. If each couple is to sit together
c. If all the men sit together on the right of all the women.


THANKS>
a) 8!
b) 4!*(2!*2!*2!*2!)
because the four couples can be arranged in 4! ways and each couple can be arranged in 2! ways

c) 4!*4!
The powers of two are bloody impolite!!
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