Combinatorial argument for the sum of the first $n$ integers.

Question

Can someone give a combinatorial argument (at least for $\binom{n+1}{2}$) for why $\binom{n+1}{2}=(n^2+n)/2$?

Answer 1

Here is a standard way to view this identity: There are $n+1$ people in a room. Everyone shakes hands with everyone else (one handshake per pair of people; nobody shakes his or her own hand). Let's count how many handshakes occurred:

Method 1: For each pair of people there is one handshake. So there are $n+1 \choose 2$ handshakes.

Method 2: Each of the $n+1$ people reports having been involved in $n$ handshakes. This gives $(n+1)\cdot n$ reported handshakes. However this counts each handshake twice (once by each of its participants). So there are $\frac{(n+1)\cdot n}{2}$ handshakes.

Method 3: Line up the people in a row. First person shakes everyone else's hand: $n$ shakes. The next person shakes hands with everyone except the first person: $n-1$ new shakes. The next person shakes hands with everyone except the first two people: $n-2$ new shakes. And so on. This gives a total of $n+(n-1)+(n-2)+\cdots + 1$ handshakes.

Comparing the three ways of viewing this situation, we have ${n+1\choose 2}=\frac{(n+1)\cdot n}{2}=1+2+3+\cdots+n$

Answer 2

let you have $n+1$ balls(with different color) you have to select 2 balls and put it in a box selecting first ball has $n+1$ ways and second ball has $n$ ways total $(n+1)\cdot n$ ways but as you are putting ball in a box it's order is not important this means your are counting twice this leads to $\binom{n+1}{n}=\frac{(n+1)n}{2}=\frac{n^2+n}{2}$.

Answer 3

Consider putting n-1 different balls in 3 bins. The stars and bars method would give the answer $ ^{(n-1) + 3 - 1} \!C_{n-1} = ^{n+1}\!C_2$ as the answer.

The other way we can distribute the balls is as follows:

• When we put all $n-1$ balls into one bin, the other two bins have total 0 objects in them. So for this case we have only 1 way to distribute the balls.
• When we put $n-2$ balls into one bin, we have one ball to put in the other two. There are two ways to distribute it among the other two bins (labelled as (number of balls in the first of the other two, number of balls in the other)) $(0,1),(1,0)$.

And so on, so when we put $n-k$ objects into one bin, there are $k+1$ ways to distribute the rest of the balls into the other two bins, $(k,0), (k-1, 1), (k-2, 2), ..., (1, k-1), (0,k)$.

Adding it all up we get: $1+2+\cdots + n = \frac{n(n+1)}{2} = \binom{n+1}{2}$ as required.

Answer 4

For any integer $n>1$, there are $n-1$ ways to write $n$ as a sum $n=a+b$ of two strictly positive integers $a,b>0$ (ignoring order). For example, $4=3+1=2+2=1+3$ makes three ways.

Let us now count the number of ways $S_3(n)$ to write $n>2$ as a sum $n=a+b+c$ of three strictly positive integers $a,b,c>0$. To do so, let us pick $m$ such that $1\leq m\leq \color{red}{n-1}$. We can write $k=n-m$ as a sum of two strictly positive integers $k=b+c$ in $k-1=n-m-1$ ways so we conclude that we can write $n=m+b+c$ ($m,b,c>0$) in the following number of ways: $$S_3(n)=\sum_{m=1}^{n-1}(n-m-1)=\sum_{k=1}^{n-1}(k-1)=\sum_{k=1}^{n-2}k.$$

Another way to count $S_3(n)$ is to consider $n$ objects labelled with $1,\; 2,\dots\;n$, pick two distinct numbers $p$ and $q$, $1\leq p<q\leq\color{red}{n-1}$ and make three groups: one group with the objects labelled from $1$ to $p$, one group with the objects labelled from $p+1$ to $q$ and the third group with the remaining objects. There are by definition $\binom{n-1}2=\frac{(n-1)(n-2)}2$ ways to do this.

We have now proven that $$S_3(n+1)=\binom n2=\frac{n(n-1)}2=\sum_{k=1}^{n-1}k.$$

Answer 5

To see combinatorially how the sum of the first $n$ integers match the number of ways to pick $2$ elements, $(a,b)$, from the list $1,2,...,(n+1)$, you could as follows:

1. Fix one element, $a$, as you first choice
2. Pick the next element $b$ to be strictly greater than $a$

If $a=1$ you have $n$ choices for $b$. If $a=(n+1)$ you have zero choices for $b$ since then it cannot be greater than $a$. Thus the number of choices for $b$ while $a$ runs from $1$ through $(n+1)$ can be summarized as $$ n+...+2+1+0 $$ which happens to be the sum of the first $n$ integers.

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As to why this should equal $(n^2+n)/2$ I think Suraj posted a perfectly simple and clear explanation for that.