If $n = 5$, then ${{\ (}^{n}C_{0})\ }^{2} + {(\ ^{n}C_{1})\ }^{2} + {(\ ^{n}C_{2})\ }^{2} + .\ldots + {(\ ^{n}C_{5})\ }^{2}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$\left( \ ^{n}C_{0} \right)^{2} + \left( \ ^{n}C_{1} \right)^{2} + \left( \ ^{n}C_{2} \right)^{2} + \ldots + \left( \ ^{n}C_{5} \right)^{2}$
$= \left( \ ^{5}C_{0} \right)^{2} + \left( \ ^{5}C_{1} \right)^{2} + \left( \ ^{5}C_{2} \right)^{2}$+$\left( \ ^{5}C_{3} \right)^{2} + \left( \ ^{5}C_{4} \right)^{2} + \left( \ ^{5}C_{5} \right)^{2}$
$= 1 + 25 + 100 + 100 + 25 + 1 = 252$

Q. If $n = 5$, then ${{\ (}^{n}C_{0})\ }^{2} + {(\ ^{n}C_{1})\ }^{2} + {(\ ^{n}C_{2})\ }^{2} + .\ldots + {(\ ^{n}C_{5})\ }^{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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