Number of 4 -digit numbers (the repetition of digits is allowed)...

Given,

We have to form four-digit number using the digits $1, 2, 3 \text{\text{ and } }$ 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 $1 - \left\{1, 1, 2\right\} \left\{\text{as}1 ,1,2 \text{\text{ and } } 5 \text{sum is divisible by}3\right\}$

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

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

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.