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