# Distribution of $N$ balls numbered from $1$ to $N$ without replacement

Same question as "Distribution of $N$ balls numbered $1$ to $N$ with replacement", but without replacement:

An urn contains $N$ balls numbered $1,2,3,...,N$.

I draw at random $n$ balls, one by one WITHOUT replacement.

Let $X$ the smallest number, the largest $Y$ and $S$ the sum of all the $n$ numbers

How to compute:

• the probability $P(X=x,Y=y)$ that $X=x$ AND $Y=y$
• the probability that $S=s$
• perhaps you'd better merge these questions – leonbloy May 11 '11 at 14:41
• @LeilaHatami And yet another useless edit to a 6 years old question. Well done. – Did Mar 20 '17 at 16:32
• @Did Sure you cannot find more productive ways of contributing to the site? – MR_BD Mar 20 '17 at 16:33

Here is the distribution of $(X,Y)$. The event $[x\le X,Y\le y]$ corresponds to subsets of size $n$ drawn from the set $[x,y]$ of size $y-x+1$. There are $\displaystyle{y-x+1\choose n}$ such subsets hence $$P(x\le X,Y\le y)=c{y-x+1\choose n},\qquad \frac1c={N\choose n}.$$ Decomposing the event $[X=x,Y=y]$ thanks to the events $[x\le X,Y\le y]$, $[x\le X,Y\le y-1]$, $[x+1\le X,Y\le y]$ and $[x+1\le X,Y\le y-1]$, one gets $$P(X=x,Y=y)=c{y-x+1\choose n}-2c{y-x\choose n}+c{y-x-1\choose n},$$ which can be simplified to $$P(X=x,Y=y)=c{y-x-1\choose n-2}.$$