# Combinatorial problem

• Oct 16th 2011, 10:12 AM
Dinkydoe
Combinatorial problem
Consider the problem :

Given $S=\left\{1,2,\cdots, 15\right\}$. 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:

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

I'm not seeing any connection yet. A little insight please? (Thinking)
• Oct 16th 2011, 10:22 AM
Plato
Re: Combinatorial problem
Quote:

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

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

There are $\binom{N+K-1}{N}$ 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 $a_2,~a_3,~\&~a_4$ that leaves eight 1's to place anywhere.
• Oct 16th 2011, 03:39 PM
Dinkydoe
Re: Combinatorial problem
That shows that the number we're looking for is ${12\choose 8}$

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