Consider all 1000-element subsets of the set $A = { 1, 2, 3, … , 2015 }$. From each such subset choose the least element. The arithmetic mean of all of these least elements is $frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
It will be:
$$frac{a_1 + a_2 + … + a_{k}}{k}$$
Obviously, one set is: ${1, 2, 3, …, 1000}$ another could be: ${1, 3, … , 1000,1001}$.
There are $2015 – 1 + 1 = 2015$ elements in set $A$. So:
$binom{2015}{1000}$ of 1000 element subsets are present.
This problem is very tough for me to solve right now…
HINTS only please! Thank you!!