Let S = (1, 2, 3, ... , 50). The number of non-empty subsets A of S such that product of element in A is even, is  $2^{\mathrm{m}}\left(2^{\mathrm{n}}-1\right)$ then the value of (m + n) is equal to

Given, Set S = {1,2,3, ... 50}.
Total number of non-empty subset of ${ }^{\text{'}}{\mathrm{S}}^{\mathrm{\prime }}=2^{50}-1$
Now, number of non-empty subset of  S in which only
odd number. {1,3,5,... 49} occurs $=2^{25}-1$
So, the required number of non-empty subsets of  S,
such that product of elements is even.
$\begin{array}{r}\left(2^{50}-1\right)-\left(2^{25}-1\right)=2^{50}-1-2^{25}+1\\ =2^{50}-2^{25}=2^{25}\left(2^{25}-1\right)\end{array}$

Here, $\mathrm{m}=\mathrm{n}=25$

$\because \mathrm{m}+\mathrm{n}=25+25=50$