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JEE Advanced 2018 · Paper 2, Mathematics question 14

Let

X=(10C1)2+2(10C2)2+3(10C3)2+⋯+10(10C10)2,X = \left({}^{10}C_1\right)^2 + 2\left({}^{10}C_2\right)^2 + 3\left({}^{10}C_3\right)^2 + \cdots + 10\left({}^{10}C_{10}\right)^2,

where 10Cr{}^{10}C_r, r∈{1,2,⋯ ,10}r \in \{1, 2, \cdots, 10\} denote binomial coefficients. Then, the value of 11430X\frac{1}{1430}X is _____ .

Answer and worked solution

Official answer: 646

Rewrite each term. Using r(10r)=10(9r−1)r\binom{10}r=10\binom9{r-1} and (10r)=(1010−r)\binom{10}r=\binom{10}{10-r}:
X=∑r=110r(10r)2=10∑r=110(9r−1)(1010−r).X=\sum_{r=1}^{10}r\binom{10}r^2=10\sum_{r=1}^{10}\binom9{r-1}\binom{10}{10-r}.

Vandermonde. The sum is the coefficient count for choosing 9 from 19:
X=10(199)=10×92378=923780.X=10\binom{19}9=10\times92378=923780.

Value.
X1430=646.\frac X{1430}=646.

Answer: 646

Topics:Properties of binomial coefficients
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