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Mathematics Question on Binomial theorem

15C0.5C5+15C1.5C4+15C2.5C3+15C3.5C2^{15}{{C}_{0}}{{.}^{5}}{{C}_{5}}{{+}^{15}}{{C}_{1}}{{.}^{5}}{{C}_{4}}{{+}^{15}}{{C}_{2}}{{.}^{5}}{{C}_{3}}{{+}^{15}}{{C}_{3}}{{.}^{5}}{{C}_{2}} +15C4.5C1{{+}^{15}}{{C}_{4}}{{.}^{5}}{{C}_{1}} is equal to

A

220−25{{2}^{20}}-{{2}^{5}}

B

20!5!15!\frac{20!}{5!15!}

C

20!5!15!−1\frac{20!}{5!15!}-1

D

20!5!15!−15!5!10!\frac{20!}{5!15!}-\frac{15!}{5!10!}

Answer

20!5!15!−15!5!10!\frac{20!}{5!15!}-\frac{15!}{5!10!}

Explanation

Solution

Now, coefficient of x15{{x}^{15}} in (1+x)20{{(1+x)}^{20}} = coefficient of x15{{x}^{15}} in (1+x)15(1+x)5{{(1+x)}^{15}}{{(1+x)}^{5}}
⇒\Rightarrow 20C15=^{20}{{C}_{15}}= coefficient of x15{{x}^{15}} in (15C0x15+15C1x14{{(}^{15}}{{C}_{0}}{{x}^{15}}{{+}^{15}}{{C}_{1}}{{x}^{14}} +15C2x13+15C3x12+15C4x11+15C5x10){{+}^{15}}{{C}_{2}}{{x}^{13}}{{+}^{15}}{{C}_{3}}{{x}^{12}}{{+}^{15}}{{C}_{4}}{{x}^{11}}{{+}^{15}}{{C}_{5}}{{x}^{10}}) (5C0x5+5C1x4+5C2x3+5C3x2{{(}^{5}}{{C}_{0}}{{x}^{5}}{{+}^{5}}{{C}_{1}}{{x}^{4}}{{+}^{5}}{{C}_{2}}{{x}^{3}}{{+}^{5}}{{C}_{3}}{{x}^{2}} +5C4x+5C5){{+}^{5}}{{C}_{4}}x{{+}^{5}}{{C}_{5}})
⇒\Rightarrow 20C15=15C0.5C5+15C1.5C4+15C2.5C3^{20}{{C}_{15}}{{=}^{15}}{{C}_{0}}{{.}^{5}}{{C}_{5}}{{+}^{15}}{{C}_{1}}{{.}^{5}}{{C}_{4}}{{+}^{15}}{{C}_{2}}{{.}^{5}}{{C}_{3}} +15C3.5C2+15C4.5C1+15C55C0{{+}^{15}}{{C}_{3}}{{.}^{5}}{{C}_{2}}{{+}^{15}}{{C}_{4}}{{.}^{5}}{{C}_{1}}{{+}^{15}}{{C}_{5}}^{5}{{C}_{0}}
⇒\Rightarrow 15C0.5C5+15C1.5C4+15C2.5C3+15C3.5C2^{15}{{C}_{0}}{{.}^{5}}{{C}_{5}}{{+}^{15}}{{C}_{1}}{{.}^{5}}{{C}_{4}}{{+}^{15}}{{C}_{2}}{{.}^{5}}{{C}_{3}}{{+}^{15}}{{C}_{3}}{{.}^{5}}{{C}_{2}} +15C4.5C1=20C15−15C55C0{{+}^{15}}{{C}_{4}}{{.}^{5}}{{C}_{1}}{{=}^{20}}{{C}_{15}}{{-}^{15}}{{C}_{5}}^{5}{{C}_{0}}
=20!5!5!−15!5!10!=\frac{20!}{5!5!}-\frac{15!}{5!10!}