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It's straightforward to show that

$$\sum_{i=1}^ni=\frac{n(n+1)}{2}={n+1\choose2}$$

but intuitively, this is hard to grasp. Should I understand this to be coincidence? Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects? What's the intuition behind this?

user238435
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  • I actually had similar encounter even with sum of an A.P Try it. It has $^{n}C_1$ and $^{n}C_2$ ! – Mann May 08 '15 at 16:17
  • What's an "A.P"? I'm sorry, I'm only a student. – user238435 May 08 '15 at 16:18
  • Arithmetic Progression. –  May 08 '15 at 16:19
  • Arithmetic progression I meant ^^ – Mann May 08 '15 at 16:19
  • This is not a coincidence. Look at my answer for a detailed combinatorial argument. – Adhvaitha May 08 '15 at 16:23
  • possible duplicate of [Proof for formula for sum of sequence $1+2+3+\ldots+n$?](http://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-ldotsn) – leonbloy May 08 '15 at 16:27
  • See also http://math.stackexchange.com/questions/44759/combinatorial-proof-that-binomial-coefficients-are-given-by-alternating-sums-of/44782#44782 – leonbloy May 08 '15 at 16:29
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    @leonbloy I'm not asking for a proof though, I was looking for intuition. – user238435 May 08 '15 at 16:30
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    @user238435 The answers include intuition-based proofs. Also see my last image in http://math.stackexchange.com/questions/44759/combinatorial-proof-that-binomial-coefficients-are-given-by-alternating-sums-of/44782#44782 – leonbloy May 08 '15 at 16:31
  • @leonbloy That thread is pretty great, so I don't know what to do now about this one, since the answers here have been very helpful. – user238435 May 08 '15 at 16:35

6 Answers6

7

Consider a tournament with $n+1$ teams each playing each other. We will count the number of matches played in two ways.

  • Every match is played between two teams. This inturn implies that the number of matches is $\dbinom{n+1}2$.
  • We will now count the number of distinct matches played team by team.
    • The number of matches played by the first team is $n$.
    • The number of matches played by the second team is $n-1$, since their match with the first team has already been accounted for.
    • The number of matches played by the third team is $n-2$, since their matches with the first and second team have already been accounted for.
    • The number of matches played by the $k^{th}$ team is $n-k+1$, since their matches with the first $k-1$ teams have already been accounted for. Hence, the total number of matches is $$n+(n-1) + (n-2) + \cdots + 1$$
Adhvaitha
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    While all the answers have been helpful, this one was pedagogically best for my grasping the intuition. Very clear explanation. – user238435 May 08 '15 at 16:37
4

Suppose that you want to choose a subset $\{m,n\}$ with two elements of the set $$ \{1,2,\dotsc,n+1\} $$ Count this in two ways one of them naturally equals $\binom {n+1}2$ and for the other observe that

If $max\{m,n\}=2$ then we have one subsets $\{m,n\}$.
If $max\{m,n\}=3$ then we have two subsets $\{m,n\}$.
$\vdots$
If $max\{m,n\}=n+1$ then we have $n$ subsets $\{m,n\}$.

Now add up these cases to derive the identity.$\square$

k1.M
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3

The intuition is that for the pairs can be listed in the following way.

$$\begin{array}{ccccccc} 1,2 & & & & & & \\ 1,3 & 2,3 & & & & & \\ 1,4 & 2,4 & 3,4 & & & & \\ 1,5 & 2,5 & 3,5 & 4,5 & & & \\ 1,6 & 2,6 & 3,6 & 4,6 & 5,6 & & \\ 1,\vdots & 2,\vdots & 3,\vdots & 4,\vdots & 5,\vdots &\ddots & \\ 1,n+1 & 2,n+1 & 3,n+1 & 4,n+1 & 5,n+1 & \cdots & n,n+1 \\ \end{array}$$

Notice that each row has length $i$ for $i=1,\ldots,n$ since the number of pairs with maximum element $i+1$ is $i$. Therefore the total number of pairs, which is $\binom{n+1}{2}$ is $\displaystyle \sum_{i=1}^n i$.

jgon
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    I doubt the person who downvoted this post will see this, but if you do, I would appreciate it in the future if you left an explanation as to why it was downvoted so that I could fix my post. – jgon May 08 '15 at 18:30
3

This is the classic proof without words, from https://maybemath.wordpress.com/

enter image description here

That doesn't help with this part of your question:

Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects?

Here's a way to rephrase @user17762 's excellent accepted answer.

Imagine $n+1$ kids in a room. Each shakes hands with all the others. Then each kid shakes hands $n$ times, so there are $n(n+1)$ handshakes - each counted twice. You can pick a pair of kids (that is, a handshake) in $n(n+1)/2$ ways. But you can also think about the kids shaking hands as they enter the room one at a time. The second kid coming has one hand to shake. The third has two, and so on, for a total of $1 + 2 + \cdots + n$.

Ethan Bolker
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If you want to choose a pair out of $n+1$ objects (for example, $\{0,1,\dots,n\}$), the possibilities are:

$\{0,1\}$, $\{0,2\}$, ..., $\{0,n\}$, giving $n$ possibilities.

$\{1,2\}$, $\{1,3\}$, ..., $\{1,n\}$ giving $n-1$ possibilities. (note that we've already picked $\{1,0\}$, so we can't repeat it here)

$\{2,3\}$, $\{2,4\}$, ..., $\{2,n\}$ giving $n-2$ possibilities.

$\ \ \ \ \vdots$

$\{n-2,n-1\}$, $\{n-2,n\}$ giving $2$ possibilities.

$\{n-1,n\}$ giving $1$ possibility.

So the number of pairs is $n+(n-1)+\dots+2+1$

Kitegi
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0

Start with $n+1$ objects, labelled $1,\dots,n+1$. We count the number of ways of choosing a pair of objects. We may always assume that the first object we choose has a lower number than the second.

Choose object number $1$. How many ways are there to choose a second object with a higher number?

Choose object number $2$. How many ways are there to choose a second object with a higher number?

Choose object number $3$. How many ways are there to choose a second object with a higher number?

$$ \dots $$

Choose object number $n$. How many ways are there to choose a second object with a higher number?

Choose object number $n+1$. How many ways are there to choose a second object with a higher number?

John Gowers
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