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 ________.

The correct answer is 206.

Step 1: Understand the Problem Statement
We are given 10 identical red pens and 14 identical blue pens, making a total of 24 pens.
We need to distribute these 24 pens among four persons, say A, B, C, and D, such that each person receives exactly 6 pens in total.

Step 2: Set up the Variable Representation
Let $r_1$, $r_2$, $r_3$, and $r_4$ be the number of red pens given to persons A, B, C, and D respectively.
Since all red pens are identical, the distribution of red pens is completely determined by non-negative integers $r_1,r_2,r_3,r_4$ such that:

$r_1+r_2+r_3+r_4=10$

Since each person must receive a total of 6 pens, the number of blue pens given to person i is automatically $6-r_{\mathrm{i}}$ for $i=1,2,3,4$.
Because the number of blue pens given to any person cannot be negative, we must have:

$0\le r_{\mathrm{i}}\le 6\text{ for each }i\in <!--\; \in \; -->\{1,2,3,4\}$

Thus, the number of ways to distribute the pens is equal to the number of non-negative integer solutions to:

$r_1+r_2+r_3+r_4=10\text{ subject to }0\le r_{\mathrm{i}}\le 6$

Step 3: Count the Total Solutions Using Generating Functions / Coefficient Finding
The number of valid distributions is the coefficient of $x^{10}$ in the expansion of:

${(1+x+x^2+x^3+x^4+x^5+x^6)}^4$

Using the finite geometric series sum formula, we can rewrite the expression as:

${\left(\frac{1-x^7}{1-x}\right)}^4={(1-x^7)}^4{(1-x)}^{-4}$

Expanding ${(1-x^7)}^4$ up to powers relevant for $x^{10}$:

${(1-x^7)}^4=1-4x^7+\text{higher terms}$

Expanding ${(1-x)}^{-4}$ using standard binomial expansion:

${(1-x)}^{-4}=\underset{k=0}{\overset{\infty }{\sum <!--\; \sum \; -->}}\left(\begin{array}{}⎛k+3⎞⎝3⎠\end{array}\right)x^k$

Now, we find the coefficient of $x^{10}$ in $(1-4x^7)\underset{k=0}{\overset{\infty }{\sum <!--\; \sum \; -->}}\left(\begin{array}{}⎛k+3⎞⎝3⎠\end{array}\right)x^k$:

$\text{Coefficient of}{x}^{10}=\left(\begin{array}{}⎛10+3⎞⎝3⎠\end{array}\right)-4\left(\begin{array}{}⎛3+3⎞⎝3⎠\end{array}\right)$

Step 4: Perform the Final Calculations
Calculate each binomial coefficient separately:

$\left(\begin{array}{}⎛13⎞⎝3⎠\end{array}\right)=\frac{13\times <!--\; \times \; -->12\times <!--\; \times \; -->11}{3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=13\times <!--\; \times \; -->2\times <!--\; \times \; -->11=286$

$\left(\begin{array}{}⎛6⎞⎝3⎠\end{array}\right)=\frac{6\times <!--\; \times \; -->5\times <!--\; \times \; -->4}{3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=20$

Subtracting the constrained cases:

$\text{Total ways}=286-4\times <!--\; \times \; -->20=286-80=206$

Thus, the number of ways to distribute the pens under the given conditions is 206.