Question

If $n$ and $r$ are integers such that $1\le r\le n$, then $n.C(n-1,r-1)$ $=$

• A. $C(n,r)$
• B. $n.C(n,r)$
• C. $rC(n,r)$
• D. $(n-1).C(n,r)$

Answer 1

The correct option is C $rC(n,r)$

$\because C(n,r)=\frac{n!}{(n-r)!r!}$

$\therefore n.C(n-1,r-1)=n\cdot \frac{(n-1)!}{(n-1-r+1)!(r-1)!}$

$\therefore n.C(n-1,r-1)$

$=n\cdot \frac{(n-1)!}{(n-r)!(r-1)!}$

$=r\cdot \frac{n!}{(n-r)!r!}$, .........................Multiplying with ${}^{\prime }{r}^{\prime }$ on both numerator and denominator & $n\cdot (n-1)!=n!$

$=r\cdot C(n,r)$