Animated Solution for Mathematics - Permutations and Combinations: The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is

To form a $5$-digit number from the set $\{1,2,3,5,6,7\}$ that is a multiple of $6$, we must satisfy two conditions simultaneously: the number must be divisible by $2$ and by $3$.

Divisibility by $2$ is a local constraint requiring the last digit to be even. Divisibility by $3$ is a global constraint requiring the sum of all digits to be a multiple of $3$.

The sum of the available digits is:

Since we are forming a $5$-digit number, we must exclude exactly one digit from the set of six. Because the total sum $24$ is divisible by $3$, the sum of the remaining five digits will be divisible by $3$ if and only if the excluded digit is also a multiple of $3$.

The multiples of $3$ in our set are $3$ and $6$. This leads us to two mutually exclusive cases.

Excluding the Digit $3$

If we exclude $3$, our set becomes $\{1,2,5,6,7\}$. The sum of these digits is $21$, which is divisible by $3$.

To satisfy the divisibility by $2$ condition, the last digit must be even. In this set, the even digits are $2$ and $6$.

We have $2$ choices for the last position. The remaining $4$ positions can be filled by the remaining $4$ digits in $4!$ ways:

Excluding the Digit $6$

If we exclude $6$, our set becomes $\{1,2,3,5,7\}$. The sum of these digits is $18$, which is divisible by $3$.

In this set, the only even digit is $2$. Therefore, we have only $1$ choice for the last position.

The remaining $4$ digits can be arranged in the first $4$ positions in $4!$ ways:

By summing the results of our two independent cases, we find the total number of valid $5$-digit numbers:

The total number of $5$-digit multiples of $6$ that can be formed from the given set is $72$.