Re: Combinatorial problem

Quote:

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.

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..?