Distributing Identical Red and Blue Pens Among Persons

To find the number of ways to distribute 10 identical red pens and 14 identical blue pens among 4 persons such that each person gets exactly 6 pens, we observe that specifying the number of red pens $r_i$ received by person $i$ uniquely determines their number of blue pens $b_i = 6 - r_i$.

Thus, the problem reduces to finding the number of integer solutions to: $r_1 + r_2 + r_3 + r_4 = 10$ subject to $0 \le r_i \le 6$ for each $i \in \{1, 2, 3, 4\}$.

Using the principle of inclusion-exclusion:

1. The total number of non-negative integer solutions without the upper bound restriction is given by: $\binom{10 + 4 - 1}{4 - 1} = \binom{13}{3} = 286$

2. We subtract the invalid solutions where at least one $r_i \ge 7$. Selecting 1 person out of 4 to have $r_i \ge 7$ leaves $10 - 7 = 3$ red pens to distribute freely among 4 persons: $\binom{4}{1} \times \binom{3 + 4 - 1}{4 - 1} = 4 \times \binom{6}{3} = 4 \times 20 = 80$

Since it is impossible for more than one person to receive 7 or more red pens (as $7 + 7 = 14 > 10$), the total number of valid distributions is: $286 - 80 = 206$