Question 11

Let $r_i$ be the number of red pens given to the $i$-th person. Since each person gets a total of 6 pens, the number of blue pens for the $i$-th person is $b_i = 6 - r_i$. Because $b_i \ge 0$, we must have $0 \le r_i \le 6$. The total number of red pens is 10, so $\sum_{i=1}^4 r_i = 10$. The number of ways is the coefficient of $x^{10}$ in $(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)^4$. $(1 + x + \dots + x^6)^4 = (\frac{1-x^7}{1-x})^4 = (1-x^7)^4 (1-x)^{-4}$. Coeff of $x^{10}$ in $(1 - 4x^7 + 6x^{14} - \dots) \sum_{k=0}^{\infty} \binom{k+3}{3} x^k$: $= 1 \times \binom{10+3}{3} - 4 \times \binom{3+3}{3} = \binom{13}{3} - 4\binom{6}{3} = 286 - 4(20) = 286 - 80 = 206$.