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If ${ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_n$ respectively are the binomial coefficients in the expansion of $(1+x)^n$, then when $n=10, \sum_{r=1}^{10}{ }^n C_r \cdot r(r-4)=$
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7680
Given $n=10$, then $\sum_{r=1}^{10}{ }^n C_r r(r-4)$
$=\sum_{r=1}^{10}{ }^n C_r r((r-1)-3)=\sum_{r=1}^{10}\left(r(r-1){ }^n C_r-3 r{ }^n C_r\right)$
$=\sum_{r=1}^{10}\left(r(r-1) \frac{n(n-1)}{r(r-1)}{ }^{n-2} C_{r-2}-3 \cdot r \frac{n}{r} \cdot{ }^{n-1} C_{r-1}\right)$
$=\sum_{r=1}^{10}\left(n(n-1){ }^{n-2} C_{r-2}-3 n \cdot{ }^{n-1} C_{r-1}\right)$
$=n(n-1) \sum_{r=1}^{10}{ }^{n-2} C_{r-2}-3 n \sum_{r=1}^{10}{ }^{n-1} C_{r-1}$
Given, $n=10$ and $\sum_{r=1}^{10}{ }^n C_r=2^n$
$\therefore 10(10-1) \cdot 2^{10-2}-3 \times 10 \cdot 2^{10-1}$
$=10 \times 9 \times 2^8-30 \times 2^9=(90-60) \cdot 2^8$
$=30 \times 256=7680$
$=\sum_{r=1}^{10}{ }^n C_r r((r-1)-3)=\sum_{r=1}^{10}\left(r(r-1){ }^n C_r-3 r{ }^n C_r\right)$
$=\sum_{r=1}^{10}\left(r(r-1) \frac{n(n-1)}{r(r-1)}{ }^{n-2} C_{r-2}-3 \cdot r \frac{n}{r} \cdot{ }^{n-1} C_{r-1}\right)$
$=\sum_{r=1}^{10}\left(n(n-1){ }^{n-2} C_{r-2}-3 n \cdot{ }^{n-1} C_{r-1}\right)$
$=n(n-1) \sum_{r=1}^{10}{ }^{n-2} C_{r-2}-3 n \sum_{r=1}^{10}{ }^{n-1} C_{r-1}$
Given, $n=10$ and $\sum_{r=1}^{10}{ }^n C_r=2^n$
$\therefore 10(10-1) \cdot 2^{10-2}-3 \times 10 \cdot 2^{10-1}$
$=10 \times 9 \times 2^8-30 \times 2^9=(90-60) \cdot 2^8$
$=30 \times 256=7680$
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