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Question: Let \( X={{\left( ^{10}{{C}_{1}} \right)}^{2}}+{{\left( ^{10}{{C}_{1}} \right)}^{2}}+3{{\left( ^{10}...

Let X=(10C1)2+(10C1)2+3(10C1)2...+10(10C1)2 X={{\left( ^{10}{{C}_{1}} \right)}^{2}}+{{\left( ^{10}{{C}_{1}} \right)}^{2}}+3{{\left( ^{10}{{C}_{1}} \right)}^{2}}...+10{{\left( ^{10}{{C}_{1}} \right)}^{2}}, where 10Cr,r=1,2,3,...,10^{10}{{C}_{r}},r=\\{1,2,3,...,10\\} denote binomial coefficients .Then find the value of 11430X\dfrac{1}{1430}X.

Explanation

Solution

Convert the given expression to generalized form to apply the standard formula of 2n−1Cn−1^{2n-1}{{C}_{n-1}}. Then make the given data a specialized case of this to find out the required value.

Complete step-by-step answer:
The given expression can written in generalized form X=∑r=0nr(nCr)2=∑r=0nr(nCr)(nCr)X={{\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}}^{2}}=\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}\left( ^{n}{{C}_{r}} \right).
We can replace nCr=nr(n−1Cr−1)^{n}{{C}_{r}}=\dfrac{n}{r}\left( ^{n-1}{{C}_{r-1}} \right) . Now the expression transforms to
\begin{aligned} & X={{\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}}^{2}} \\\ & \Rightarrow X=\sum\limits_{r=0}^{n}{r}\left( ^{n}{{C}_{r}} \right)\left( ^{n}{{C}_{r}} \right) \\\ & \Rightarrow X=\sum\limits_{r=0}^{n}{r\left( ^{n}{{C}_{r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right)\left( \dfrac{n}{r} \right) \\\ & \Rightarrow X=n\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right) \\\ \end{aligned}$$$$$ We use the fact that ^{n}{{C}{r}}{{=}^{n}}{{C}{n-r}} We also know from theory of binomial expansion that $\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{n-r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right){{=}^{2n-1}}{{C}_{n-1}}$. Putting it in above equation $$$$ $\begin{aligned} & X=n\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right) \\\ & \Rightarrow X=n\sum\limits_{r=0}^{n}{\left( ^{n}{{C}_{n-r}} \right)}\left( ^{n-1}{{C}_{r-1}} \right) \\\ & \Rightarrow X=n\left( ^{2n-1}{{C}_{n-1}} \right) \\\ \end{aligned}
Now we apply for the special case as asked in the question . So
X=10⋅(20−1C10−1)=10(19C10)X=10\cdot \left( ^{20-1}{{C}_{10-1}} \right)=10\left( ^{19}{{C}_{10}} \right)

We have been asked to find out the value of 11430X\dfrac{1}{1430}X. So we first factorize 1430 as 1430=10.11.131430=10.11.13. Using this obtained value to substitute in the above equation.
11430X=1019×18×...119×8×...2×10×11×13=646\dfrac{1}{1430}X=10\dfrac{19\times 18\times ...11}{9\times 8\times ...2\times 10\times 11\times 13}=646
The required value is 646.$$$$

Note: We need to be careful of wrong substitution as it may lead to incorrect results. We need to be also careful of the fact that the question is asking the values of 11430X\dfrac{1}{1430}X not XX. So do not end the solution at XX .