1

I am having trouble proving this identity using combinatoric or algebraic proof. As someone pointed me out it is somehow related to pascals triangle recurrence.

$$\sum_{i=0}^k \binom{n+i}{i} = \binom{n+k+1}{k}$$

I found a question where this equation was posted but didnt understand any answers there... Combinatorial proofs - how?

Could someone help me out?

Community
  • 1
VOid
  • 83
  • So if you have $15$ names, you can EITHER seat the first $12$ on the jury and send the other three home, OR seat the first $13$ and choose one to be an alternate who will fill in if one of the $12$ falls ill, OR seat the first $14$ and choose two of those as alternates, OR seat the first $15$ and choose three of those as alternates. And the total number of ways to do that is the same as the number of ways to choose $3$ out of $16$. And the question is then: what are the $16$ things and what are the $3$ that are chosen that somehow determine who are the $12$ jurors and who are the alternates? – Michael Hardy Oct 06 '14 at 18:36
  • See http://math.stackexchange.com/questions/489191/combinatorics-identity-question/489211#489211 – user84413 Oct 06 '14 at 20:37

1 Answers1

0

So if you have 15 names, you can EITHER seat the first 12 on the jury and send the other three home, OR seat the first 13 and choose one to be an alternate who will fill in if one of the 12 falls ill, OR seat the first 14 and choose two of those as alternates, OR seat the first 15 and choose three of those as alternates. And the total number of ways to do that is the same as the number of ways to choose 3 out of 16. And the question is then: what are the 16 things and what are the 3 that are chosen that somehow determine who are the 12 jurors and who are the alternates? $$ \binom{12}0 + \binom{13}1 + \binom{14}2 + \binom{15}3 = \binom{16}3 $$ This is the same as $$ \binom{12}{12}+ \binom{13}{12} + \binom{14}{12} + \binom{15}{12} = \binom{16}3 $$ Here's why this works: You have your list of $15$ names plus the one dummy name. You will choose three of the $16$ names.

If the dummy is not included among those three, then those three are the alternates and you have $\dbinom{15}{3}$ ways that can happen.

If the dummy is included among the three chosen, then there are two others. If the $15$th actual name is not included among those two, the those are the two alternates. There are $\dbinom{14}{2}$ ways that can happen.

If the dummy and the $15$th name are among the three chosen, then there is one other. If that one other is not the $14$th actual name, than that one is the alternate. There are $\dbinom{13}1$ ways that can happen.

If the dummy and the $15$th and $14$th names among those chosen, then there are no alternates. There are $\dbinom{12}0$ ways that can happen.

Thus the $15$th person is an alternate only if the dummy is not chosen and the $15$th person is. The $14$th person is an alternate only if the dummy and the $15$th person are not chosen and the $14$th person is.

Michael Hardy
  • 1