Question

$ ^{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

• A. $ {{2}^{20}}-{{2}^{5}} $
• B. $ \frac{20!}{5!15!} $
• C. $ \frac{20!}{5!15!}-1 $
• D. $ \frac{20!}{5!15!}-\frac{15!}{5!10!} $

Answer 1

The Correct Option is D

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!} $