The total number of 5 -digit numbers, formed by using the digits 1,...

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\}$</p><p>Arrangement will be$____2 \text{and } ____6</p><p>In unit place2&6$can take place because for multiple of$6number should be even & sum of digits should be multiple of3$.</p><p>So total way in which remaining number can be taken will be$4 ! \times 2 = 48$ways</p><p>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 $= 24 + 48 = 72$