Mathematics - Permutation Combination Question with Solution | TestHub

Given set $\left\{1,2,3,5,6,7\right\}$

Here sum is $1+2+3+5+6+7$ $=24$

So there will be two cases

Case (i) When numbers are $\left\{1,2,5,6,7\right\}$

Arrangement will be $\_ \_ \_ \_ 2 \& \_ \_ \_ \_ 6$

In unit place $2$ & $6$ can take place because for multiple of $6$ number should be even & sum of digits should be multiple of $3$.

So total way in which remaining number can be taken will be $4!\times 2$ $=48$ ways

Case (ii) When numbers are  $\left\{1,2,3,5,7\right\}$

Arrangement will be $\_ \_ \_ \_ 2$ 

So total way in which remaining number can be taken will be $4!=24$ ways

Adding case (i) & case (ii), we get

required $5$-digit numbers $=24+48=72$