Browse Questions

The sum of the digits in unit place of all the numbers formed with the help of 3,4,5 and 6 taken all at a time is

$\begin{array}{1 1}(A)\;432\\(B)\;108\\(C)\;36\\(D)\;18\end{array}$

Sum of the digits in the units place
$\Rightarrow (4-1)! (3+4+5+6)$
$\Rightarrow 3!(3+4+5+6)$
$\Rightarrow 6\times 18$
$\Rightarrow 108$
Hence (B) is the correct answer.