JEE Main 2025 Apr 2 Shift 1, Mathematics Q3: Combinations

A 10-term sequence with exactly five 1s and exactly three 2s must also contain $10-5-3=2$ zeros. So we are arranging the multiset $\{1,1,1,1,1,2,2,2,0,0\}$.

The number of distinct arrangements is given by the multinomial coefficient:

$$\frac{10!}{5!3!2!}=\frac{3628800}{120\times 6\times 2}=\frac{3628800}{1440}=2520$$

$\therefore$ Option (3) is correct.