JEE Main 2026 — Mathematics Algebra

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

If player $A$A wins the series in exactly $n$n games, then $A$A must win the $n$n-th game, and $A$A must win exactly $4$4 out of the first $n-1$n−1 games. The number of ways this can happen is given by $^{n-1}C_{4}$n−1C4​.

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

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

Total ways $= ^{4}C_{4} + ^{5}C_{4} + ^{6}C_{4} + ^{7}C_{4} + ^{8}C_{4}$=4C4​+5C4​+6C4​+7C4​+8C4​

Using the property $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$nCr​+nCr−1​=n+1Cr​, we can simplify the sum:

$$^{4}C_{4} = ^{5}C_{5}$$4C4​=5C5​

$$^{5}C_{5} + ^{5}C_{4} = ^{6}C_{5}$$5C5​+5C4​=6C5​

$$^{6}C_{5} + ^{6}C_{4} = ^{7}C_{5}$$6C5​+6C4​=7C5​

$$^{7}C_{5} + ^{7}C_{4} = ^{8}C_{5}$$7C5​+7C4​=8C5​

$$^{8}C_{5} + ^{8}C_{4} = ^{9}C_{5}$$8C5​+8C4​=9C5​

Therefore, the total number of ways is:

$$^{9}C_{5} = \dfrac{9!}{5!4!} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$9C5​=5!4!9!​=4×3×2×19×8×7×6​=126

Answer: $126$126