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

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 :\

4!/2! + 4!/2! - 3!2!2!