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  4. 15C0.5C5+15C1.5C4+15C2.5C3+15C3.5C2 +15C4.5C1 is equal to

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Q. $ ^{15}{{C}_{0}}{{.}^{5}}{{C}_{5}}{{+}^{15}}{{C}_{1}}{{.}^{5}}{{C}_{4}}{{+}^{15}}{{C}_{2}}{{.}^{5}}{{C}_{3}}{{+}^{15}}{{C}_{3}}{{.}^{5}}{{C}_{2}} $ $ {{+}^{15}}{{C}_{4}}{{.}^{5}}{{C}_{1}} $ is equal to

KEAMKEAM 2009Binomial Theorem

Solution:

Now, coefficient of $ {{x}^{15}} $ in $ {{(1+x)}^{20}}
$= coefficient of $ {{x}^{15}} $ in $ {{(1+x)}^{15}}{{(1+x)}^{5}} $
$ \Rightarrow $ $ ^{20}{{C}_{15}}= $ coefficient of $ {{x}^{15}} $ in $ {{(}^{15}}{{C}_{0}}{{x}^{15}}{{+}^{15}}{{C}_{1}}{{x}^{14}} $ $ {{+}^{15}}{{C}_{2}}{{x}^{13}}{{+}^{15}}{{C}_{3}}{{x}^{12}}{{+}^{15}}{{C}_{4}}{{x}^{11}}{{+}^{15}}{{C}_{5}}{{x}^{10}}) $ $ {{(}^{5}}{{C}_{0}}{{x}^{5}}{{+}^{5}}{{C}_{1}}{{x}^{4}}{{+}^{5}}{{C}_{2}}{{x}^{3}}{{+}^{5}}{{C}_{3}}{{x}^{2}} $ $ {{+}^{5}}{{C}_{4}}x{{+}^{5}}{{C}_{5}}) $
$ \Rightarrow $ $ ^{20}{{C}_{15}}{{=}^{15}}{{C}_{0}}{{.}^{5}}{{C}_{5}}{{+}^{15}}{{C}_{1}}{{.}^{5}}{{C}_{4}}{{+}^{15}}{{C}_{2}}{{.}^{5}}{{C}_{3}} $ $ {{+}^{15}}{{C}_{3}}{{.}^{5}}{{C}_{2}}{{+}^{15}}{{C}_{4}}{{.}^{5}}{{C}_{1}}{{+}^{15}}{{C}_{5}}^{5}{{C}_{0}} $
$ \Rightarrow $ $ ^{15}{{C}_{0}}{{.}^{5}}{{C}_{5}}{{+}^{15}}{{C}_{1}}{{.}^{5}}{{C}_{4}}{{+}^{15}}{{C}_{2}}{{.}^{5}}{{C}_{3}}{{+}^{15}}{{C}_{3}}{{.}^{5}}{{C}_{2}} $ $ {{+}^{15}}{{C}_{4}}{{.}^{5}}{{C}_{1}}{{=}^{20}}{{C}_{15}}{{-}^{15}}{{C}_{5}}^{5}{{C}_{0}} $
$=\frac{20!}{5!5!}-\frac{15!}{5!10!} $