Let  $A_1=\left\{1\right\},A_2=\{1,2\};A_3=\{1,2,3\};\mathrm{......}{A}_r=\{1,2,3,\mathrm{......},r\}$ and so on and let  $S=\{1,2,3,4,5,6\}$. Define $P\left(i\right)=$   Number of subsets of  S  having exactly one element of  $A_i$ , Then the value of  $\sum _{i=1}^{6}P\left(i\right)$

$i=1,P\left(1\right)=$number of subsets of S which is 1 there
                  = total subsets of S- subsets which donot have 1 $=2^6-2^5=32$
                   
$i=2,P\left(2\right)=$subsets which have out of  $\{1,2\}$
say 1 is there and 2 not there  subsets of  $\{1,3,4,5,6\}$  in which 1 is there  $=2^5-2^4=16$
for the other case 2 is there and 1 not there  $=2^5-2^4=16$   $\therefore P\left(2\right)=2\left(16\right)=32$
similarly  $P\left(3\right)=3(2^4-2^3)=24$ ,  $P\left(4\right)=4(2^3-2^2)=16$,   $P\left(5\right)=5(2^2-2)=10$ and

$$\begin{gathered} P\left(6\right)=6 \\ \therefore \sum _{i=1}^{6}P\left(i\right)=32+32+24+16+10+6=120 \end{gathered}$$