JEE Main 2026 — Permutation Combination Question with Solution

Player $A$ wins the series if $A$ wins $5$ games before $B$ wins $5$ games. The series can last for a minimum of $5$ games and a maximum of $9$ games.

If player $A$ wins the series in exactly $n$ games, then $A$ must win the $n$-th game, and $A$ must win exactly $4$ out of the first $n-1$ games. The number of ways this can happen is given by ${}^{n-1}{C}_4$.

The possible values for $n$ are $5,6,7,8,$ and $9$.

Total number of ways for $A$ to win the series is the sum of the number of ways $A$ can win in $5,6,7,8,$ or $9$ games:

Total ways ${=}^4{C}_4{+}^5{C}_4{+}^6{C}_4{+}^7{C}_4{+}^8{C}_4$

Using the property ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$, we can simplify the sum:
${}^4{C}_4{=}^5{C}_5$
${}^5{C}_5{+}^5{C}_4{=}^6{C}_5$
${}^6{C}_5{+}^6{C}_4{=}^7{C}_5$
${}^7{C}_5{+}^7{C}_4{=}^8{C}_5$
${}^8{C}_5{+}^8{C}_4{=}^9{C}_5$

Therefore, the total number of ways is:
${}^9{C}_5=\frac{9!}{5!4!}=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}=126$

Answer: $126$