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  4. If 18 C 15+2( 18 C 16)+ 17 C 16+1= n C 3 then n is equal to

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Q. If ${ }^{18} C _{15}+2\left({ }^{18} C _{16}\right)+{ }^{17} C _{16}+1={ }^{ n } C _3$ then $n$ is equal to

Binomial Theorem

Solution:

${ }^{18}{ }^{18} C _{15}+2\left({ }^{18} C _{16}\right)+{ }^{17} C _{16}+1={ }^{ n } C _3$
$\Rightarrow{ }^{18} C _{15}+{ }^{18} C _{16}+{ }^{18} C _{16}+{ }^{17} C _{16}+{ }^{17} C _{17}={ }^n C _3 \left[\because{ }^{ n } C _{ n }=1\right] $
$\Rightarrow{ }^{18+1} C _{16}+{ }^{18} C _{16}+{ }^{17+1} C _{17}={ }^n C _3 \left[\because{ }^{ n } C _{ r -1}+{ }^{ n } C _{ r }={ }^{ n +1} C _{ r }\right] $
$\Rightarrow{ }^{19} C _{16}{ }^{16}{ }^{18} C _{16}{ }^{16}+{ }^{18} C _{17}={ }^{17} C _3={ }^{19} C _{16}+{ }^{18+1} C _{17}={ }^n C _3 $
$\Rightarrow{ }^{19} C _{16}+{ }^{19} C _{17}={ }^{ n } C _3={ }^{19+1} C _{17}={ }^{ n } C _3 $
$\Rightarrow{ }^{20} C _3={ }^{ n } C _3 \left[\because{ }^{ n } C _{ n - r }={ }^{ n } C _{ r }\right] $
$\therefore n =20$