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I was looking through practice questions and need some guidance/assistance in Fermat's combinatorial identity. I read through this on the stack exchange, but the question was modified in the latest edition of my book.

The book asks for a combinatorial argument (no computations needed) to establish the identity:

$${n \choose k} = \sum_{i=k}^n {i-1 \choose k-1} \text{ where }n\ge k$$

Hint given: Consider set of numbers $1$ through $n$, and how many subsets of size $k$ have $i$ as their highest numbered member?

I'm still getting a grasp on the way these arguments work, so any help is greatly appreciated.

Community
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sahimat
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  • Related:https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1?noredirect=1&lq=1. – StubbornAtom Feb 19 '18 at 17:48

1 Answers1

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Here's the first part to get you started. Fix $i \in \{1, \ldots, n\}$. To choose a subset of size $k$ with largest element $i$, we choose $i$, and then we must choose the remaining $k-1$ elements from $\{1, 2, \ldots, i-1\}$. (If we choose an element in the range $\{i+1, i+2, \ldots, n\}$, then $i$ won't be the largest element!)

Can you see where the summation comes from? What are the possible values for the largest element of a $k$-element subset of $\{1, 2, \ldots, n\}$?

Edit. Editing in my comment from below:

We are going to count the number of $k$-element subsets of $\{1,\ldots,n\}$, of which there are $\binom{n}{k}$, ordering them by their largest element. To choose a subset of size $k$ with largest element $i$, we choose $i$, and then we must choose the remaining $k-1$ elements from $\{1,2,\ldots,i−1\}$, which yields $\binom{i-1}{k-1}$ possibilities. Since this is a $k$-element subset, the largest element $i$ must be in the range $\{k,k+1,\ldots,n\}$, so we sum over this range.

Viktor Vaughn
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    Yes, this helps a lot actually. By picking one number smaller at a time, it covers the whole summation, and thus gives me the total n choose k (whatever k is defined as). Thank you for the insight. (I think that's what you were getting at). //side note: I would upvote, but lack the "rep", but will return when I can – sahimat Sep 18 '14 at 02:40
  • I honestly don't see this pattern –  Feb 11 '16 at 11:22
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    We are going to count the number of $k$-element subsets of ${1, \ldots, n}$, of which there are $\binom{n}{k}$, in order of their largest element. To choose a subset of size $k$ with largest element $i$, we choose $i$, and then we must choose the remaining $k−1$ elements from ${1,2,\ldots,i−1}$, which yields $\binom{i-1}{k-1}$ possibilities. Since this is a $k$-element subset, the largest element $i$ must be in the range ${k, k+1, \ldots, n}$, so we sum over this range. – Viktor Vaughn Feb 11 '16 at 14:39
  • @Quasicoherent why would we need those largest elements? – parvin Aug 05 '17 at 17:13
  • @parvin I don't understand your question. Could you clarify? Every subset has a largest element, and we are simply organizing our counting by grouping together those subsets with the same largest element. – Viktor Vaughn Aug 05 '17 at 17:55
  • I see, you mean you consider one indivisual which is the largest, each time, then calculation the rest's combinations? so every element would be the biggest for one time ? – parvin Aug 07 '17 at 08:18
  • @parvin Yes, I think that's right. You could try computing an example, e.g., enumerating all subsets of $3$ elements from ${1,2,3,4,5}$ to try it out. – Viktor Vaughn Aug 07 '17 at 22:42
  • I understand this but why are we summing over all the values and how does this help us establish equality? –  Jul 03 '20 at 19:52
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    @user1234 Did you read my comment above? I've edited it into the answer now. I think this answers your question. – Viktor Vaughn Jul 03 '20 at 20:58