For $r = 0, 1, 2, \dots, 10,$ 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 $n_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$
is equal to

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$B_{10} - C_{10}$

$A_{10} (B_{10}^{2} - C_{10} A_{10})$

$C_{10} - B_{10}$

answer is D.

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Detailed Solution

We have

$A_{r} =^{10} C_{r}, B_{r} =^{20} C_{r}$ and $C_{r} =^{30} C_{r}$

$\begin{array}{l} ∴ \sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r}) \\ = B_{10} \sum_{r = 1}^{10} A_{r} B_{r} - C_{10} \sum_{r = 1}^{10} {(A_{r})}^{2} \\ = B_{10} (\sum_{r = 1}^{10}^{10} {C_{r}}^{20} C_{r}) - C_{10} \sum_{r = 1}^{10} {(^{10} C_{r})}^{2} \\ = B_{10} (\sum_{r = 1}^{10}^{10} {C_{r}}^{20} C_{20 - r}) - C_{10} \{\sum_{r = 0}^{10} {(^{10} C_{r})}^{2} - 1\} \end{array}$

$= B_{10} \{\sum_{r = 0}^{10}^{10} {C_{r}}^{20} C_{20 - r} - 1\} - C_{10} \{\sum_{r = 0}^{10} {(^{10} C_{r})}^{2} - 1\}$

$\begin{array}{l} = B_{10} \times \end{array}$ {Coefficient of $x^{20}$ in $(1 + x)^{10} (x + 1)^{20} - 1\}$

$- C_{10} \{^{20} C_{10} - 1\}$

$\begin{array}{l} = B_{10} \times \{^{30} C_{20} - 1\} - C_{10} \{^{20} C_{10} - 1\} \\ = B_{10} \{^{30} C_{10} - 1\} - C_{10} \{^{20} C_{10} - 1\} \\ = B_{10} (C_{10} - 1) - C_{10} (B_{10} - 1) = C_{10} - B_{10} \end{array}$