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16Cn +16C9 =17C9
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\[16Cn \implies \frac{16!}{n!(16-n)!}\] \[16C9 \implies \frac{16!}{9!(16-9)!} \] \[17C9 \implies \frac{17!}{9!(17-9)!}\] does that help?
so how do you find n?
u know that property: \[^nC_r+^nC_{r-1}=^{n+1}C_r\] ?
@virtus use this property, and put n=16,r=9 so now can u find your 'n' ??
oh i see, your job is to find \(n\) for thie equality correct?
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