Counting question? Hi could anyone do this question for me or something similar cause I got a lot questions like this  and I cant solve them thanks
Let $n \geq 66$ be an integer and consider the set $S = \{1,2,\ldots,n\}$. 
Let $k$ be an integer with $66 \leq k \leq n$. How many 
      $66$-element subsets of $S$ are there whose largest element is 
      equal to $k$? 
Use the result in the first part to prove that 
$ \sum_{k=66}^n {{k-1} \choose {65}} = {n \choose 66}$
 A: To make a $66$-element set whose largest entry is $k$, we must choose $65$ numbers from the numbers $1$ to $k-1$ to keep $k$ company. There are $\binom{k-1}{65}$ ways to do this. 
Now let us count the number of $66$-element subsets of $\{1,2,\dots,n\}$ in two different ways.
First of all, the number is clearly $\binom{n}{66}$.
Secondly, these subsets are of various types: Largest element is $66$; largest element is $67$; largest element is $68$; and so on up to largest element is $n$.
By the first paragraph, the number of subsets $66$-element subsets with largest element $66$ is $\binom{65}{65}$. The number of $66$-element subsetssubsets with largest element $67$ is $\binom{66}{65}$. The number of $66$-element subsets with largest element $68$ is $\binom{67}{65}$. and so on, until the number of $66$-element subsets with largest element $n$ is $\binom{n-1}{65}$. Thus the total number of $66$-element subsets is
$$\binom{65}{65}+\binom{66}{65}+\binom{67}{65}+\cdots +\binom{n-1}{65}.\tag{1}$$
This yields the desired result. We have counted the number of $66$-element subsets, one way simply as $\binom{n}{66}$, and also the fancy way represented by (1). The two answers must be the same.   
A: For the purpose of finding subgroups whose largest element is k, elements larger than k might as well not exist. The largest number, k, is fixed, but we need 65 more, all of which are less than k. Thus the answer is ${k-1 \choose 65}$. ${n \choose 66}$ is the number of subgroups of 66 elements with n as the largest integer. Since $n$ is in the subgroup, one of the 66 elements is fixed and the others are not. Thus we are interested in finding 65 element subgroups to add $n$ to to make 66 element subgroups. Each of these 65 element subgroups have some largest element between 65 and n-1. Add up all the ways to find such groups across all k and you have the answer.
