JEE Mathematics 2025 Question with Solution

Given: Seven-digit numbers are formed using only the digits $1$, $2$, and $3$, and the sum of the digits is $11$.

Find: The number of such numbers in the set $P$.

Let $n_1, n_2, n_3$ denote the number of times digits $1, 2, 3$ occur respectively. Then,

$$n_1+n_2+n_3=7$$

and

$$n_1+2n_2+3n_3=11$$

Using $n_1=7-n_2-n_3$ in the second equation,

$$(7-n_2-n_3)+2n_2+3n_3=11$$ $$7+n_2+2n_3=11$$ $$n_2+2n_3=4$$

Now find all non-negative integer solutions.

1. If $n_3=0$, then $n_2=4$ and hence $n_1=3$. Number of arrangements:

$$\frac{7!}{3!4!0!}=35$$

2. If $n_3=1$, then $n_2=2$ and hence $n_1=4$. Number of arrangements:

$$\frac{7!}{4!2!1!}=105$$

3. If $n_3=2$, then $n_2=0$ and hence $n_1=5$. Number of arrangements:

$$\frac{7!}{5!0!2!}=21$$

If $n_3>2$, then $n_2$ becomes negative, so no more cases are possible. Therefore, total number of elements is

$$35+105+21=161$$

Hence, the number of elements in the set $P$ is $161$. The correct option is C.