JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Permutations and Combinations

The captain is one fixed batsman and the vice-captain one fixed bowler, so two of the ten places are already filled and eight more players must be chosen from the remaining $6$ batsmen and $5$ bowlers. A valid team of ten carries at least $4$ batsmen and at least $4$ bowlers, which permits only the splits $4$ batsmen with $6$ bowlers, $5$ batsmen with $5$ bowlers, and $6$ batsmen with $4$ bowlers. Since one batsman and one bowler are already in, the further selections are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)\left(\genfrac{}{}{0ex}{}{5}{5}\right)=20$, $\left(\genfrac{}{}{0ex}{}{6}{4}\right)\left(\genfrac{}{}{0ex}{}{5}{4}\right)=75$ and $\left(\genfrac{}{}{0ex}{}{6}{5}\right)\left(\genfrac{}{}{0ex}{}{5}{3}\right)=60$ respectively. The total number of selections is $20+75+60=155$.