Using mathematical induction, verify:

$${n \choose k} = \frac{n}{k} {n-1 \choose k-1}$$

For the base case, I need to have k=0? Or both n and k? When just k=0, there is a fraction with 0 as the denominator? The LHS is = 1, but I don't think this makes the RHS = 1...


  • $\begingroup$ Use en.wikipedia.org/wiki/Pascal's_rule with fix $k>0$ $\endgroup$ Feb 25, 2015 at 4:08
  • $\begingroup$ Induction is a fairly silly way to prove this identity; are you required to use it? $\endgroup$ Feb 25, 2015 at 4:10
  • $\begingroup$ ${n \choose k} = \frac{n}{k} {n-1 \choose k-1}$ iff $k{n \choose k} = n {n-1 \choose k-1}$ $\endgroup$
    – BigM
    Feb 25, 2015 at 4:29
  • $\begingroup$ Yes, it is required. $\endgroup$ Feb 25, 2015 at 13:07

1 Answer 1


$$\frac nk{n-1\choose k-1}=\frac nk\frac{(n-1)!}{((n-1)-(k-1))!(k-1)!}=\frac{n!}{(n-k)!k!}={n\choose k}.$$


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