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Q.1.     If ^{n}C_{8}=^{n}\; \! \! \! \! C_{2}, find ^{n}C_{2}

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Given : ^{n}C_{8}=^{n}\; \! \! \! \! C_{2},

We know that ^nC_a=^nC_b  \Rightarrow a=b\, \, \, or\, \, n=a+b

^{n}C_{8}=^{n}\; \! \! \! \! C_{2},

\Rightarrow n=8+2

\Rightarrow n=10

 

^{n}C_{2}=^{10}C_2

       =\frac{10!}{(10-2)!2!}

        =\frac{10!}{8!2!}

        =\frac{10\times 9\times 8!}{8!2!}

       =5\times 9=45

Thus the answer is 45

 

 

 

Posted by

seema garhwal

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