4

Possible Duplicate:
simple binomial theorem proof

Why is $${6\choose 0} + {7\choose 1} + \ldots + {n+6 \choose n} = {n+7 \choose n}\;?$$

Community
  • 1
Pam
  • 43
  • Write out ${n\choose k}$ for each term on the left-hand side and then combine. – JohnD Dec 14 '12 at 04:42
  • What do you mean by combine? I thought you could only combine things with the same top number? – Pam Dec 14 '12 at 04:42
  • Add up all of those binomial coefficients on the left-hand side. – JohnD Dec 14 '12 at 04:43
  • 6!/0!6! + 7!/1!6! + ... + (n+6)!/n!*6! = 1 + 7 + 28 + ... – Pam Dec 14 '12 at 04:45
  • Hint for the algebraic argument: write $\binom{6}{0}$ as $\binom{7}{0}$ – Isomorphism Dec 14 '12 at 04:56
  • Hint for the combinatorial argument: Say you $n + 7$ balls numbered from $1$ to $n+7$ in a bag. What is the number of ways of picking $n$ balls from the bag, if you have decided to pick balls numbered $1$ to $i-1$ but not the $i$th ball? – Isomorphism Dec 14 '12 at 05:09
  • @Pam did you also answer this question: http://math.stackexchange.com/a/262660/19341 with a different account? – draks ... Dec 20 '12 at 16:07

2 Answers2

3

HINTS: The algebraic argument is a proof by induction: verify the equality for $n=0$, and show that if $${6\choose 0} + {7\choose 1} + \ldots + {n+6 \choose n} = {n+7 \choose n}\;,$$ then $${6\choose 0} + {7\choose 1} + \ldots + {(n+1)+6 \choose {n+1}} = {(n+1)+7 \choose {n+1}}\;.$$ This is a pretty straightforward application of the Pascal’s triangle identity.

For the combinatorial argument, observe that $\binom{n+7}n=\binom{n+7}7$ is the number of ways to choose a $7$-element subset of $\{1,2,\dots,n+7\}$, and $\binom{k+6}k=\binom{k+6}6$ is the number of ways to choose a $7$-element subset of $\{1,\dots,n+7\}$ whose largest element is the number $k+1$. That is, to choose a $7$-element subset whose largest element is $10$, you first choose $10$, and then you have to choose $6$ of the numbers $\{1,\dots,9\}$.

Brian M. Scott
  • 616,228
1

The standard recursive formula for binomial coefficients is

$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

There is a technique for recursion called "unrolling," where you repeatedly substitute a recurrence into its equation for itself. Here if we reverse the terms and then substitute repeatedly we get:

$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$

$\binom{n}{k} = \binom{n-1}{k} + \binom{n-2}{k-1} + \binom{n-2}{k-2}$

$\vdots$

$\binom{n}{k} = \binom{n-1}{k} + \binom{n-2}{k-1} + \binom{n-3}{k-2} + \cdots + \binom{n-k}{0}$

William Macrae
  • 743
  • However, unrolling is not actually a proof; it’s one of the ways that one might discover the formula, and it suggests but does not actually incorporate the proof by induction. – Brian M. Scott Dec 14 '12 at 05:01
  • This is true. The question said "show," and my assumption was that either the author did not need a proof or if she did she would have an idea of how to translate this into a proof. As you've pointed out, this should be done by induction. – William Macrae Dec 14 '12 at 05:05
  • 2
    Show almost always means prove. – Brian M. Scott Dec 14 '12 at 05:05
  • 1
    Although usually a "show" proof is less formal than a "prove" proof, in my experience. – Joe Z. Dec 14 '12 at 05:31
  • @Joe: Some people do use the terms that way, but you can’t count on it: many of us use them completely synonymously. – Brian M. Scott Dec 14 '12 at 06:42