JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

Let r_i and b_i be the number of red and blue pens given to the i-th person, where i = 1, 2, 3, 4. Given that each person receives exactly 6 pens, we have: r_i + b_i = 6 b_i = 6 - r_i Since the number of blue pens must be non-negative (b_i 0), we get r_i 6. Also, r_i 0. The total number of red pens is 10, so: r_1 + r_2 + r_3 + r_4 = 10 The number of ways to distribute the pens is the number of non-negative integer solutions to this equation subject to 0 r_i 6. This is equivalent to finding the coefficient of x^ 10 in the expansion of (1 + x + x^2 + + x^6)^4. (1 + x + x^2 + + x^6)^4 = ( 1 - x^7 1 - x )^4 = (1 - x^7)^4 (1 - x)^ -4 Expanding both terms: (1 - x^7)^4 = 1 - 4x^7 + 6x^ 14 - (1 - x)^ -4 = _ k=0 ^ ^ k+3 C_ 3 x^k We need the coefficient of x^ 10 in the product (1 - 4x^7 + ) _ k=0 ^ ^ k+3 C_ 3 x^k. Coefficient of x^ 10 = 1 ( coefficient of x^ 10 in (1-x)^ -4 ) - 4 ( coefficient of x^3 in (1-x)^ -4 ) = ^ 10+3 C_ 3 - 4 ^ 3+3 C_ 3 = ^ 13 C_ 3 - 4 ^ 6 C_ 3 = 13 12 11 3 2 1 - 4 6 5 4 3 2 1 = 286 - 4 20 = 286 - 80 = 206 Answer: 206