No of ways choosing $6$ persons from $6$ Married couples Find number of ways to choose $6$ persons from $6$ married couples such that there is exactly one couple.
To solve, I have first chosen one couple which is $6$ ways.
Now remaining $4$ persons have to chosen from $10$ persons left such that among these four there should be no couple. Number of ways this can be done in 
$$\binom{10}{4}-5=205$$
so the total number of ways is $205 \times 6=1230$,
but answer is given $480$. What is my mistake?
 A: Pick the couple: $\binom 61 =6$ ways
Pick the other couples that you're going to choose one partner from: $\binom 54 = 5$
Pick which partner from those couples: $2^4=16$
Answer : $6\cdot 5\cdot 16 = 480$
Your mistake is in overestimating the number of ways to pick the four without including a couple.  As you can see here there are only $5\cdot 16=80$ ways to do that.

You can also get the $80$ options of picking $4$ people without any couples from $5$ couples by inclusion-exclusion, starting with your initial "free" choice of $4$:
$$\binom {10}{4} - \binom {5}{1}\binom {8}{2} + \binom {5}{2}\binom {6}{0} = 210 - 5\cdot 28 + 10\cdot 1 = 80$$
where the first subtraction is all the ways of choosing one couple and any two other people, and the final addition remedies the "oversubtraction" of cases where two couples were chosen.
A: Let married couples as $aa',bb',cc',dd',ee',ff'$
Then number of ways of choosing exactly one married couple is $\displaystyle \binom{6}{1} = 6$
now choosing $4$ other people which are not couple , is $$\displaystyle \binom{5}{4}\times \binom{2}{1}\times \binom{2}{1}\times \binom{2}{1}\times \binom{2}{1} = 5 \times 16$$
So total number of ways is $\displaystyle 6\times 5 \times 16 = 480$
