# Finding the number of combinations of two dates from n dates where one precedes the other?

Suppose that I am given n consecutive dates for buying and selling shares . So what are the number of ways of choosing a pair of buy date and sell date such that buy date always precede sale date ?

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Pick any two distinct dates. There are $\binom{n}{2}=\frac{n(n-1)}{2}$ ways to do this. Label the earlier one buy, the later one sell.
I have interpreted "earlier" as meaning different. It is always possible to buy one day and sell later the same day. With that interpretation, there are $n$ additional possibilities.
This is a binomial coefficient. One high school notation for the same thing is ${}_n\text{C}_2$, often pronounced in english "$n$ choose $2$." There are other notations, like $C(n,2)$, or $C_2^n$. To see that there are $\frac{n(n-1)}{2}$ ways to do this, write down one date. There are $n$ ways to do this. For each choice, there are $n-1$ ways to write a different date, for a total of $n(n-1)$. For half these choices, the first date chosen will be later than the second date chosen.So the number of pairs $(a,b)$ such that $a$ is earlier than $b$ is $\frac{n(n-1)}{2}$. – André Nicolas Aug 1 '12 at 7:43