Please explain why $1 + 2 + 3 + \dotsb + n = \binom{n+1}{2}$ like I am a 5 year old

Question

We know $1 + 2 + 3 + \dotsb + n = \frac{(n+1)n}{2}$, and $\binom{n+1}{2}=\frac{(n+1)n}{2}$ . How could a summation equal a combination? Please give us a simple explanation.

Answer 1

Well, the smart aleck answer would be because $1 + 2 + ... + n = \frac {n(n+1)}2$[1] and also ${n+1 \choose 2} = \frac {n(n+1)}2$[2]. Although it's not that smart alecky as the math does work and is easy to verify.

But as to why this could have been expected:

If you choose two numbers for $n$. There are $n$ choices what the larger number is. And if the larger number was $k$ there are $k-1 $ choices for what the smaller number could be.

That is. If large number is:

$2$ there is $1$ possible value for the smaller number:$1$

$3$ there are $2$ possible values for the smaller number: $1$ or $2$.

$4$ there are $3$ possible values for the smaller number: $1$, $2$ or $3$.

....

$n$ there are $n-1$ possible value for the smaller number: $1$, $2$, $3$,.....,$n-1$.

$n+1$ there are $n$ possible values for the smaller number: $1$, $2$, $3$,......,$n-1$, or $n$.

Add those up and there are $1 + 2 + 3 +...... + n$ ways to chose the same two numbers.

[1] $M = 1 + 2 + ...... + n$

$M = n + (n-1) + ....... + 1$

$M+M = (n+1) + (n-1 + 2) + .......+ (2 + n-1) + (1 + n)$

$2M= (n+1) + (n+1) + ..... + (n+1) + (n+1)= n(n+1)$

$M = \frac {n(n+1)}2$.

[2] ${n+1 \choose 2} = \frac {(n+1)!}{2!((n+1) - 2)!}$

$= \frac {1*2*3*..... *(n-1)*n*(n+1)}{(1*2)(1*2*3*......*(n-1)}=$

$\frac {n(n+1)}{2}$.

Answer 2

Yes. Suppose you want to choose 2 from $n+1$ objects which are numbered as $1, 2, \ldots, n+1$. To complete the task, you need two objects no. $j$ and no. $k$ where $j \neq k$. WLOG, assume $j <k$ Then you could first pick an object numbered $k$, then you pick a no. $j$ where $k \geqslant j+1 $. You have $k-1$ choose to pick no. $j$ for each $k=2,3,4,\ldots, n+1$. Thus the number of choices in total is $\sum_{k=2}^{n+1} (k-1) =1 +2+ \cdots + n$. Hence the equation.

Answer 3

Let we have $n+1$ balls tagged from $1$ to $n+1$. The number of cases in which you can draw $2$ distinct balls out from the set without replacement is $\binom{n+1}{2}$. Alternatively you can do this as following $$\text{the number of cases of drawing out }(1,i)\text{ when }i=2,3,\cdots ,n,n+1=n\\\text{the number of cases of drawing out }(2,i)\text{ when }i=3,\cdots ,n,n+1=n-1\\\text{the number of cases of drawing out }(3,i)\text{ when }i=4,\cdots ,n,n+1=n-2\\.\\.\\.\\\text{the number of cases of drawing out }(n-1,i)\text{ when }i=n,n+1=2\\\text{the number of cases of drawing out }(n,i)\text{ when }i=n+1=1\\\text{total number of cases}=1+2+3+\cdots+n$$which yields to $$1+2+3+\cdots+n=\binom{n+1}{2}$$

Answer 4

Say there are $n+1$ people in a line

$$p_1p_2\dots p_np_{n+1}$$

Now define a rule to choose a pair without repetition: pick $p_i$ and find $p_j$ but notice $i<j$. This will guarantee that no repetition occurs

When $i=1,$ $(n+1)-1$ people remains so there are $n$ ways to choose (s)he's mate.
When $i=2,$ $(n+1)-2$ people remains so there are $n-1$ ways to choose (s)he's mate.

So clearly $n+(n-1)+\dots+1$ ways to choose a pair from $n+1$ people.

This observation is from double factorial notation: $!!$,
for example when there are $n$ people, where $2\vert n$ then use the same rule but this time once a pair is chosen remove them from the list so $(n-1)(n-3)\cdot\dots\cdot(1)$ ways to divide them into pairs.

Answer 5

One more attempt:

1) $\binom{n+1}{2}$ is the number of ways of choosing $2$ objects out of $(n+1)$ distinct objects.

2) Consider $(1,2,3,....., n, n+1)$ the $(n+1)$ -tuple representing the $n+1$ object.

Choose $n+1$ and pair it with $1,2,...(n-1),n$ :

$n$ pairs.

Choose $n$ and pair it with $1,2,...., (n-1)$,

$(n-1)$ pairs.

Choose $(n+1)-k$ , $0 \le k \le n-1$.

and pair with

$1,2,......((n+1)-(k+1))$.

$((n+1)-(k+1))= n-k$ pairs.

::::::::::: :::::::::::

Choose $((n+1)- k)$, with $k=n-1$:

$(n+1) -(n-1)=2$

and pair it with 1.

$1$ pair.

Altogether, the number of pairs:

$n +(n-1)+(n-2)+(n-3)+.......+ 3 + 2+1.$