JEE Mathematics 2025 Question with Solution

Given: A sequence has $10$ terms, each term is either $0$, $1$, or $2$. It contains exactly five $1$'s and exactly three $2$'s.

Find: The number of distinct such sequences and hence the correct option.

This is a problem of permutations with repetitions.

The remaining terms must be zeros, so the number of $0$'s is

$$10 - 5 - 3 = 2$$

Now arrange $5$ ones, $3$ twos, and $2$ zeros in $10$ positions.

Using combinations:

$$\binom{10}{5} = 252$$

Choose positions for the five $1$'s.

From the remaining $5$ positions, choose positions for the three $2$'s:

$$\binom{5}{3} = 10$$

The remaining $2$ positions are automatically occupied by $0$'s.

Therefore, total number of sequences is

$$\binom{10}{5}\binom{5}{3} = 252 \times 10 = 2520$$

Equivalently,

$$\frac{10!}{5!3!2!} = 2520$$

Therefore, the total number of such sequences is $2520$. The correct option is C.

Given: There are $10$ positions with repeated entries: five $1$'s, three $2$'s, and two $0$'s.

Find: The number of distinct sequences.

Since identical symbols repeat, use the multinomial form directly:

$$\text{Number of sequences} = \frac{10!}{5!3!2!}$$

Evaluating,

$$\frac{10!}{5!3!2!} = 2520$$

This shortcut works because dividing by $5!$, $3!$, and $2!$ removes overcounting caused by interchanging identical $1$'s, $2$'s, and $0$'s.

Therefore, the correct option is C.