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From 10 married couples, we want to select group of 6 that is not allowed to contain a married couple. How many choices are there?

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There are 10 married couples

Total number of men = 10
Total number of women = 10

We need to select group of 6 out of these 20 people that is not allowed to contain a married couple.

This can be done in 3 ways

1) All men  = $\binom{10}{6} = 210$

2) All women = $\binom{10}{6} = 210$

3) Both men and women

     1 men and 5 women = $\binom{10}{1} . \binom{9}{5}$ = 1260 (for every men we can select 5 women out of 9 women because 1 is his wife)
     2 men and 4 women = $\binom{10}{2} . \binom{8}{4}$ = 3150 (for every 2 men we can select 4 women out of 8 women)
     3 men and 3 women = $\binom{10}{3} . \binom{7}{3}$ = 4200 (for every 3 men we can select 3 women out of 7 women)
     4 men and 2 women = $\binom{10}{4} . \binom{6}{2}$ = 3150 (for every 4 men we can select 2 women out of 6 women)
     5 men and 1 women = $\binom{10}{5} . \binom{5}{1}$ = 1260 (for every 5 men we can select 1 women out of 5 women)

Total combinations = 210 + 210 + 1260 +3150 + 4200 + 3150 +1260 =  $13440$ combinations
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Total ways of selecting 6 people out of a group of 20 = 20C6 = 38760

But we have counted the case when we have either one couple, two couples or three couples. We will subtract these cases to get the desired result.

No of ways of selecting exactly one couple = 10C1 * 18 * 16 * 14 * 12 / 4! = 20160

(Reason, select a pair in 10C1 ways, now to select four more, there are 18 ways to select next person, after that we can't choose his/her partner, so for the third person we only have 16 ways, and so on 14 and 12, but here we have considered all permutations of 4 people, so to eliminate that we divide by 4!). Similarly for the other cases below,

No of ways of selecting exactly two couples = 10C2 * 16 * 14 / 2! = 5040

No of ways of selecting exactly three couples = 10C3 = 120

Answer  = 38760 - 20160 - 5040 - 120 = 13440

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