For r=0, 1, ldots, 10,let A_{r}, B_{r} and C_ | vedclass

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(D) Given $A_r = \binom{10}{r}$,$B_r = \binom{20}{r}$,$C_r = \binom{30}{r}$.We need to evaluate $S = \sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r) = B_

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$C_{10}-B_{10}$

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(D) Given $A_r = \binom{10}{r}$,$B_r = \binom{20}{r}$,$C_r = \binom{30}{r}$.We need to evaluate $S = \sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r) = B_{10} \sum_{r=1}^{10} A_r B_r - C_{10} \sum_{r=1}^{10} A_r^2$.Using the property $\sum_{r=0}^{n} \binom{n}{r} \binom{m}{k-r} = \binom{n+m}{k}$,we have:$\sum_{r=0}^{10} A_r B_r = \sum_{r=0}^{10} \binom{10}{r} \binom{20}{r} = \sum_{r=0}^{10} \binom{10}{10-r} \binom{20}{r} = \binom{30}{10} = C_{10}$.Since $A_0 = 1$ and $B_0 = 1$,$\sum_{r=1}^{10} A_r B_r = C_{10} - A_0 B_0 = C_{10} - 1$.Similarly,$\sum_{r=0}^{10} A_r^2 = \sum_{r=0}^{10} \binom{10}{r} \binom{10}{10-r} = \binom{20}{10} = B_{10}$.Since $A_0 = 1$,$\sum_{r=1}^{10} A_r^2 = B_{10} - A_0^2 = B_{10} - 1$.Substituting these into $S$:$S = B_{10}(C_{10} - 1) - C_{10}(B_{10} - 1) = B_{10} C_{10} - B_{10} - C_{10} B_{10} + C_{10} = C_{10} - B_{10}$.

For $r=0, 1, \ldots, 10$,let $A_{r}, B_{r}$ and $C_{r}$ denote,respectively,the coefficient of $x^{r}$ in the expansions of $(1+x)^{10}$,$(1+x)^{20}$ and $(1+x)^{30}$. Then $\sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r)$ is equal to

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