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# Four married couples have bought 8 seats in a row for a

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Joined: 12 Dec 2004
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Four married couples have bought 8 seats in a row for a [#permalink]  06 Jan 2005, 08:59
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Question Stats:

100% (01:01) correct 0% (00:00) wrong based on 0 sessions
Four married couples have bought 8 seats in a row for a football game.

a) In how many ways can they be seated if each couple is to sit together with the husband to the left of his wife?
b) In how many ways can they be seated if each couple is to sit together
c) In how many ways can they be seated if all the men are to sit together and all the women are to sit together?
d) In how many ways can they be seated if none of the men are to sit together and none of the women are to sit together?

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a) 24 ways = First position should be a man so 4 ways second can be 1 way (his wife) so = 4x1x3x1x2x1x1x1 = 24 ways

b) First position can be selected 8 ways and second one only 1 (either man or wife), 3rd can be selected from 6 remaining ways and 4th only 1 way (same logic as above)...etc. So number of ways = 8x1x6x1x4x1x2x1 = 384 ways

c) 1152 ways ===> 2(4x4x3x3x2x2x1x1)
Intern
Joined: 12 Dec 2004
Posts: 33
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Quote:
a) 24 ways = First position should be a man so 4 ways second can be 1 way (his wife) so = 4x1x3x1x2x1x1x1 = 24 ways

b) First position can be selected 8 ways and second one only 1 (either man or wife), 3rd can be selected from 6 remaining ways and 4th only 1 way (same logic as above)...etc. So number of ways = 8x1x6x1x4x1x2x1 = 384 ways

c) 1152 ways ===> 2(4x4x3x3x2x2x1x1)

Thoese are all valid

for d, the equation becomes

4! ways of arranging the men
4! ways of arranging the women
2 ways a man and woman can be arranged between themselves

4!*4!*2 = 1152

Last edited by schumacher on 07 Jan 2005, 07:22, edited 2 times in total.
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