Probability 4 different numbers $ a, b, c , d$ are solution of $a+b=c+d$ Let $N=\{1,\ldots,n\}$ We choose $a$, $b$, $c$, $d$ - different random numbers from $N$. What is probability of $a+b=c+d$?
 A: First note that, for any fixed $k\in\{2,\dotsc,2n\}$, we have
$$ \mathbb P(a+b=k)
= \sum_{i=1}^{k-1} \mathbb P(a=i) \mathbb P(b=k-i)
= \sum_{i=1}^{k-1} \frac1n\cdot\frac1n
= \frac{k-1}{n^2}
$$
Thus
$$ \mathbb P(a+b=c+d)
= \sum_{k=2}^{2n} \mathbb P(a+b=k) \mathbb P(c+d=k)
= \sum_{k=2}^{2n} \frac{(k-1)^2}{n^4}
= \frac1{n^4} \sum_{k=1}^{2n-1} k^2
$$
Perhaps you can take it from here.
A: Suppose $a<b<c<d$.
Let the common sum be S, and suppose $S\le n+1$.  I will do $S>n+1$ later.
There are $P(S)=\lfloor(S-1)/2\rfloor$ possible pairs $a,b$.  We need two pairs with the same $S$.  That can be done in $P(S)(P(S)-1)/2$ ways.  The total number of ways is 
$$\sum_{S=5}^{n+1}P(S)(P(S)-1)/2$$
This depends on whether $n$ is even or odd.
The number of ways with $S>n+1$ is the same as for $2n+2-S$, so you have $$\sum_{S=5}^{n}P(S)(P(S)-1)/2$$
Finally, divide by the total number of foursomes, which is $_nC_4$
A: Looking at a number line, the top of your keyboard for added convenience, helps to visualize this problem.
Considering all sums of distinct pairs from $\{1,2,3,4,...,n\}$ we can see that a sum, $S$, of $5$ is the smallest case where the event in question can happen (namely $(a,b,c,d) = (1,4,2,3)$).  We can go through all possible sums by keeping $a$ fixed at $1$ and incrementing $b$ by $1$ to see what remaining pairs in between satisfy the equality.  The number of pairs is just: 
$$\frac{(b-1)-1}{2}$$ 
when $b$ is even ($S$ is odd) and:
$$\frac{(b-1)-2}{2}$$ 
when $b$ is odd ($S$ is even).
To get the result we only need to sum the number of pairs across even $b$ and odd $b$ and divide by $\binom{n}{4}$.  We also need to consider even $n$ and odd $n$ separately.
For even $n$ up to $S=n+1$:
Number of pairs when $b$ is even:  $\sum_{k=2}^{n/2}\frac{2k-2}{2}=\sum_{k=2}^{n/2}(k-1)$
Number of pairs when $b$ is odd: $\sum_{k=2}^{n/2-1}\frac{2k+1-3}{2}=\sum_{k=2}^{n/2}\frac{2k-2}{2}=\sum_{k=2}^{n/2-1}(k-1)$
In total their sum is $2\sum_{k=2}^{n/2}(k-1)-(n/2-1)=2\sum_{k=1}^{n/2-1}k-(n/2-1)$
$=n/2(n/2-1)-(n/2-1)=(n/2-1)^2$
Starting this process over again with $a=2, b=5$ then $a=3,b=6$, etc. and counting the desired pairs is done similarly and accounts for all sums. 
The approach for odd $n$ is similar. 
