Given n couples means n husband and n wife.
Let i = number of husband coming to party
Then along with him wife should also accompany.
So remaining number of wife = n – i
So, remaining n-i wife has 2 choices, either she comes to party or don’t.
${\binom{n}{i}}$ = Choosing i husband out of n that come to party
$2^{n-i}$ = 2 choices for remaining n-i wife
So equation reduces to : ( Take summation over all i's)
= $\sum_{1}^{n}$ ${\binom{n}{i}}$ * $2^{n-i}$
= $\sum_{1}^{n}$ ${\binom{n}{i}}$ * ($\frac{2^{n}}{2^{i}}$ )
= $2^{n}$ $\sum_{1}^{n}$ ${\binom{n}{i}}$ / ($2^{i}$ )
= $2^{n}$ ( $\frac{{\binom{n}{0}}} {2^{0}} $ + $\frac{{\binom{n}{1}}} {2^{1}}$ + $\frac{{\binom{n}{2}}} {2^{2}}$ …….. + $\frac{{\binom{n}{n}}} {2^{n}})$)
= $2^{n}$ $\frac{ ( {\binom{n}{0}} * (2^{n}) + {\binom{n}{1}} * (2^{n-1}) + {\binom{n}{2}} * (2^{n-2}) + .……… + {\binom{n}{n}} * (2^{0}) )}{2^{n}} $
Using Binomial Identity:
$(a+b)^{n}$ = $\sum_{1}^{n} \binom{n}{i} a^{i} b^{n-i}$
Put a =1 and b = 2, we get
$3^{n}$ = ( $ {\binom{n}{0}} * (2^{n}) + {\binom{n}{1}} * (2^{n-1}) + {\binom{n}{2}} * (2^{n-2}) + .……… + {\binom{n}{n}} * (2^{0}) )$
Finally putting value of above identity into equation,
= $2^{n}$ $\frac{ ( {\binom{n}{0}} * (2^{n}) + {\binom{n}{1}} * (2^{n-1}) + {\binom{n}{2}} * (2^{n-2}) + .……… + {\binom{n}{n}} * (2^{0}) )}{2^{n}} $
= $2^{n}$ ($ \frac{3^{n}}{2^{n}}) $
= $3^{n}$