Number of 4 -digit numbers (the repetition of digits is allowed) which are made using the digits 1 ,  …

Given,

We have to form four-digit number using the digits $1, 2, 3 \& 5$

Now for number to be divisible by $15$ we need to fix the last digit as $5$ as  $\_ \_ \_ 5$, so that number can be divisible by $5$ and for divisibility by $3$ the addition of all number should be divisible by $3$,

So making cases where sum is divisible by $3$ we get,

Case $1- \left\{1,1,2\right\}$ $\left\{\text{as }1,1,2 \& 5\text{ sum is divisible by }3\right\}$

So, total ways for case $1$ is $3$ ways,

Case $2-\left\{5,1,1\right\}\to 3 \text{ways}, \left\{3,3,1\right\}\to 3 \text{ways} \& \left\{3,2,2\right\}\to 3\text{ ways}$

So, total ways for case $2$ is $9$ ways.

Case $3- \left\{5,3,2\right\}\to 6\text{ ways}$

Case $4- \left\{5,5,3\right\}\to 3 \text{ways}$

So, adding all cases we get, $3+9+6+3=21$ ways.