What is the intuition behind
$$ \binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} $$
? I can't grasp why picking a group of $k$ out of $n$ bijects to first picking a group of $k-1$ out of $n-1$ and then a group of $k$ out of $n-1$.
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http://math.stackexchange.com/questions/86093/why-is-n1-choose-k-n-choose-k-n-choose-k-1 – lab bhattacharjee May 15 '13 at 16:49
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@labbhattacharjee Why link to a question that has been closed as duplicate when mentioning that a question is a duplicate? – Tobias Kildetoft May 15 '13 at 16:52
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1@TobiasKildetoft, could have supplied the one not closed as duplicate, but the closed one also has some additional solutions as well – lab bhattacharjee May 15 '13 at 16:55
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https://en.wikipedia.org/wiki/Pascal%27s_triangle may be a useful visual aid as consider any spot in the triangle and notice that unless it is a 1 on the side, it is the sum of the two values above it. – JB King May 15 '13 at 18:25
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1It is not "first picking... and then..." but "either picking... or...". Remember "or" (if exclusive) leads to "$+$" while "and" leads to "$\times$". – Marc van Leeuwen May 16 '13 at 09:14
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Perhaps the question @Amzoti links to will help the OP understand, but I doubt this. The OP has a specific problem with the neat combinatorial proof, which I do not think is addressed by any of the answers (in either of the two linked questions). Perhaps the OP can clarify if these threads help him? Until then, this question should not be closed as a duplicate as it is not a duplicate! – user1729 May 16 '13 at 13:08
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@user1729: The top answer to http://math.stackexchange.com/q/20475/30222 is precisely the combinatorial, intuitive proof. – tomasz May 16 '13 at 15:40
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Yes, but my interpretation is that he doesn't "get" this proof, and wants another explanation. Maybe I am wrong. Thus, I think it would have been pertinent to ask the OP if one of those links answered his question before closing it. If so, fine. Otherwise, poor OP! – user1729 May 16 '13 at 15:44
2 Answers
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We have a group of $n$ people, one of whom is John. We want to pick a committee of $k$ people. By definition this can be done in $\binom{n}{k}$ ways.
There are $\binom{n-1}{k}$ committees of $k$ that don't include John, for we can choose any $k$ of the people other than John. And there are $\binom{n-1}{k-1}$ committees of $k$ that do include John, for we can choose any $k-1$ people to join John.
Note that automatically a committee that doesn't include John is different from a committee that includes John. So we have divided the $\binom{n}{k}$ possible committees into two groups, one of which has $\binom{n-1}{k}$ elements, and the other of which has $\binom{n-1}{k-1}$ elements.
André Nicolas
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6Great, now how do we solve the problem if nobody on the committee is named John? ;) – Michael Joyce May 15 '13 at 17:01
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6It will take a while, but we can apply for a legal change of name, and then we can use the above solution. – André Nicolas May 15 '13 at 17:03
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1@MarkAdler: Implicitly, the Change of Name Lemma above solves that problem. – André Nicolas May 15 '13 at 17:24
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You don't pick a group of $k-1$ out of $n-1$ and a group of $k$ out of $n-1$; this would correspond to multiplication. You should do one or the other.
The key is to pick one of your $n$ objects as being special. Then any collection of $k$ objects from the $n$ either consists of $k-1$ objects from the $n-1$ non-special objects, plus the special one, or it consists entirely of $k$ non-special objects. There are $\binom{n-1}{k}$ collections of $k$ non-special objects, and $\binom{n-1}{k-1}$ collections of $k$ objects including the special one.
mdp
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This answer is basically a wordier version of André's, who was first, but I decided to post it anyway for the first paragraph. – mdp May 15 '13 at 16:52