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Two couples and one single person are seated at random in a

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by FinanceBioE » Sun Sep 13, 2009 12:36 pm
Hi

New to the forum I hope this helps

Number of possible arrangements of AA BB and C
5!/2!2! = 30

Then I found how many ways you can arrange them to get couples sitting next to each other

AACBB
AABBC
AABCB

BAACB
BAABC
CAABB

BCAAB
BBAAC
CBAAB

BBCAA
BCBAA
CBBAA

4*3 = 12*2 = 24 since two pairs of couples

1- 24/30 = 6/30

Sorry if its wrong :\
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by prindaroy » Sun Sep 13, 2009 2:51 pm
4!/2! + 4!/2! - 3!2!2!
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