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Show that $\binom{n}{3} = \binom{n-1}{2} + \binom{n-1}{3}$

My solution

Formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

$LHS = \binom{n}{3} = \frac{n!}{3*2(n-3)!} = \frac{n(n-1)!}{3*2(n-3)(n-4)!} = $

$ = \frac{(n-1)!}{3*2(n-4)!}*\frac{n}{n-3} = \binom{n-1}{3} * \frac{n}{n-3} $

So i got one term right. Anyone know how i could continue from here? :)

Daniel Andersson
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2 Answers2

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$\binom n3 = \binom{n-1}{2}+\binom{n-1}{3}\\ \frac{n!}{6(n-3)!} = \frac{(n-1)!}{2(n-3)!}+\frac{(n-1)!}{6(n-4)!}\\ \frac{n(n-1)(n-2)}{6} = \frac{(n-1)(n-2)}{2}+\frac{(n-1)(n-2)(n-3)}{6}\\ n = 3 + n - 3\\ n=n$ Hence proved. In the second last step, I multiplied by $\frac{6}{(n-1)(n-2)}$

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    Hey Daniel, please upvote and/or accept my answer by clicking on the tick mark to the left of my answer. It really helps out a lot man... –  May 08 '19 at 18:09
  • Thank you sir! I dont understand that last part. How is $\frac{n(n-1)(n-2)}{6} $ equal to $ \frac{(n-1)(n-2)}{2} +\frac{(n-1)(n-2)(n-3)}{6}$ – Daniel Andersson May 08 '19 at 18:18
  • Think of the expansion of the factorials in its true product form in the numerator and denominator and then see how many terms cancel out. –  May 08 '19 at 18:21
  • For example, in the LHS, $(n-3)!$ in the denominator is nothing but a factor of$n!$ in the numerator. –  May 08 '19 at 18:23
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    Thank you so much for up voting and accepting my answer –  May 08 '19 at 18:25
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Hint: Prove that $$\frac{n(n-1)(n-2)}{6}=\frac{(n-1)(n-2)}{2}+\frac{(n-1)(n-2)(n -3)}{6}$$

Dr. Sonnhard Graubner
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