Evaluate Sum of Coefficients in Binomial Expansions

To evaluate the given sum, we first recall the binomial coefficients given by the problem:

• $A_r = \binom{10}{r}$ for $r = 0, 1, \dots, 10$
• $B_r = \binom{20}{r}$ for $r = 0, 1, \dots, 20$
• $C_r = \binom{30}{r}$ for $r = 0, 1, \dots, 30$

The expression to evaluate is:
$S = \sum_{r=1}^{10} A_r (B_{10} B_r - C_{10} A_r)$

We can split the sum into two individual summations:
$S = B_{10} \sum_{r=1}^{10} A_r B_r - C_{10} \sum_{r=1}^{10} A_r^2$

Now, let us calculate each summation separately using Cauchy-Vandermonde's identity:

1. First Summation ($\sum_{r=1}^{10} A_r B_r$):
   Using the property $\binom{10}{r} = \binom{10}{10-r}$, we can write:
   $\sum_{r=0}^{10} A_r B_r = \sum_{r=0}^{10} \binom{10}{10-r} \binom{20}{r}$

This summation is equal to the coefficient of $x^{10}$ in the product $(1+x)^{10} (1+x)^{20} = (1+x)^{30}$.
Therefore:
$\sum_{r=0}^{10} A_r B_r = \binom{30}{10} = C_{10}$

Since $A_0 B_0 = \binom{10}{0} \binom{20}{0} = 1$, we have:
$\sum_{r=1}^{10} A_r B_r = \sum_{r=0}^{10} A_r B_r - A_0 B_0 = C_{10} - 1$

1. Second Summation ($\sum_{r=1}^{10} A_r^2$):
   Similarly, using $\binom{10}{r} = \binom{10}{10-r}$:
   $\sum_{r=0}^{10} A_r^2 = \sum_{r=0}^{10} \binom{10}{10-r} \binom{10}{r}$

This summation is equal to the coefficient of $x^{10}$ in $(1+x)^{10} (1+x)^{10} = (1+x)^{20}$.
Therefore:
$\sum_{r=0}^{10} A_r^2 = \binom{20}{10} = B_{10}$

Since $A_0^2 = \binom{10}{0}^2 = 1$, we have:
$\sum_{r=1}^{10} A_r^2 = \sum_{r=0}^{10} A_r^2 - A_0^2 = B_{10} - 1$

1. Substituting back into the original expression:
   $S = B_{10} (C_{10} - 1) - C_{10} (B_{10} - 1)$
   $S = B_{10} C_{10} - B_{10} - C_{10} B_{10} + C_{10}$
   $S = C_{10} - B_{10}$

Thus, the given sum is equal to $C_{10} - B_{10}$, which corresponds to Option D.