JEE Advanced 2026 Paper 1, Mathematics Section 3 Q3: Combinations

JEE Advanced2026Paper 1Mathematics Section 3

The number of ways to distribute $10$ identical red pens and $14$ identical blue pens among four persons such that each person gets $6$ pens, is _____________.

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Solution

Let $r_{i}$ and $b_{i}$ denote the number of red and blue pens given to person $i$ , for $i = 1, 2, 3, 4$ . The constraints are $r_{i} + b_{i} = 6$ for each $i$ , $\sum r_{i} = 10$ , and $\sum b_{i} = 14$ .

The second constraint follows automatically from $r_{i} + b_{i} = 6$ and $\sum r_{i} = 10$ , since $\sum b_{i} = 24 - 10 = 14$ .

Since $b_{i} = 6 - r_{i} \geq 0$ , we need $0 \leq r_{i} \leq 6$ .

Reformulation: Count non-negative integer solutions of $r_{1} + r_{2} + r_{3} + r_{4} = 10$ with each $r_{i} \leq 6$ .

By inclusion-exclusion, the unrestricted count is $(3 13) = 286$ . Subtract solutions where some $r_{i} \geq 7$ (set $r_{i}^{'} = r_{i} - 7 \geq 0$ , leaving $r_{i}^{'} + others = 3$ ): the number per offending variable is $(3 6) = 20$ , and at most one variable can violate the bound at a time (since $7 + 7 > 10$ ).

Total $= (3 13) - (1 4) (3 6) = 286 - 80 = 206$ .

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JEE Advanced 2026 Paper 1, Mathematics Section 3 Q3: Combinations

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