1. Combinatorial problem

Consider the problem :

Given $\displaystyle S=\left\{1,2,\cdots, 15\right\}$. How many subsets S' of 4 different elements can we make, such S' contain no consecutive numbers?

I like to show that this problem is equivalent with finding the number of solutions of:

$\displaystyle a_1+a_2+a_3+a_4+a_5=14; \ a_1,a_5\geq 0; \ a_2,a_3,a_4\geq 2$

I'm not seeing any connection yet. A little insight please?

2. Re: Combinatorial problem

Originally Posted by Dinkydoe
I like to show that this problem is equivalent with finding the number of solutions of:

$\displaystyle a_1+a_2+a_3+a_4+a_5=14; \ a_1,a_5\geq 0; \ a_2,a_3,a_4\geq 2$
There are $\displaystyle \binom{N+K-1}{N}$ ways to place N identical objects into K distinct cells.
Here we have fourteen identical 1's and five distinct variables.
If we go ahead and put two of the ones into each of $\displaystyle a_2,~a_3,~\&~a_4$ that leaves eight 1's to place anywhere.

3. Re: Combinatorial problem

That shows that the number we're looking for is $\displaystyle {12\choose 8}$

But I don't see how these 2 problems are equivalent. How do they relate..?