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It is well-known that $1+2+3+4+...+n= \frac{n(n+1)}{2}$, this formula can be found using simple arithmetic progression. But the sum also can be found by using ${n+1 \choose 2}$, which is n+1 choose 2. Can anyone explain the relationship between the binomial coefficient and the sum of consecutive integers? I found the binomial coefficient formula at wikipedia: http://en.wikipedia.org/wiki/Triangular_number

Dave Clifford
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4 Answers4

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If all $n + 1$ people at a party shake hands with each other such that every two people shake hands exactly once, how many handshakes are there?

Let's count the number of handshakes in two different ways:

  1. The first person shakes hands with the remaining $n$ people. The second person shook hands with the first already, so he shakes hands with the remaining $n - 1$ people. And continuing in this manner, the last-but-one person shakes hands with the last $1$ person. The last person shakes hands with none. So the number of handshakes is $n + (n - 1) + \ldots + 1$ = $1 + 2 + \ldots + n$.
  2. The number of handshakes is the number of ways of selecting a pair of people from $n + 1$ people (because every possible pair shakes hands exactly once), that is $\binom {n + 1} 2$.

Since the number of handshakes is the same no matter how you count it (correctly) $$1 + 2 + \ldots n = \binom {n + 1} 2$$

M. Vinay
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  • This is basically the handshaking lemma or degree sum formula (https://en.wikipedia.org/wiki/Handshaking_lemma) applied to the complete graph on $n$ vertices. – M. Vinay Jun 07 '14 at 07:51
  • This is very cool! Great post! I like your view on counting more than mine, more intuitive! – DanZimm Jun 07 '14 at 07:53
  • @DanZimm Thanks. Can't help thinking that way, I'm a graph theorist :P – M. Vinay Jun 07 '14 at 07:54
  • And a very wise one at that, I've seen you answering posts recently and you seem to have a vast spectrum of knowledge :D – DanZimm Jun 07 '14 at 07:56
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I think this is a better way that justifies the name Triangular Number. Consider the (right) triangular arrangement with $n$ rows and columns, the rows indexed by $i = 1, 2, \ldots n$, and the columns indexed by $0, 1, \ldots n - 1$: $$\begin{matrix}1 \\ 2 \\ 3 \\ \vdots \\ n\\ \ \end{matrix}\left| \begin{matrix} *&\\ *& *\\ *& *& *&\\ \vdots& \vdots& \vdots& \ddots\\ *& *& *& \ldots & *\\ \hline 0 & 1 & 2 & \ldots & n-1 \end{matrix}\right.$$

The total number of objects is obviously $1 + 2 + \ldots + n$ (the sum of the number of objects in each row). However, observe that each object corresponds to a unique pair of the form $(i, j)$ with $j < i$, where $i$ is the row and $j$ is the column occupied by the object. Thus the number of objects is the number of such pairs, which is the number of distinct pairs of the $n + 1$ numbers $0, 1, \ldots, n$, and that is $\binom {n + 1} 2$.

M. Vinay
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When you look at $n+1$ objects then lets try to manually count how many ways we can choose $2$ objects.

First we can count the $n$ pairs of consecutive objects, then then $n-1$ pairs that have one object in between and so on until we have the $1$ pair with $n-1$ objects in between. Thus here we can see that the number of ways to choose two objects is $n + n-1 + n-2 + \cdots + 2 + 1$ which is exactly what $\sum_{k=1}^n k$ is. Thus we come to the conclusion that $\sum_{k=1}^n k = \binom{n+1}{k}$.

DanZimm
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HINT:

If you start choosing $2$ elements from $n+1,$ say $a_i(1\le I\le n+1)$

first choose $1$ element (say $a_0$), the other can be chosen from the rest $n:\{a_1,a_2,\cdots,a_{n-1},a_n\}$ elements

for the next $a_1$ element, the other can be chosen from the rest $n-2:\{a_2,\cdots,a_{n-1},a_n\}$ elements as $a_0a_1$ has already been chosen as the order immaterial here

So on

lab bhattacharjee
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