How many solutions are there of the equation a+b+c=12, where a,b,c are non-negative integers?

We can think of any solution of this equation as a string of twelve 1's and two 0's. For example, one such solution could be a=6, b=2 and c=4. In this case we would write the solution as 11111101101111. Written in this way, all solutions of the equation must consist of fourteen symbols; exactly twelve 1's and two 0's. The only difference between solutions is the choice of where to put the two 0's. Thus the problem reduces to finding out how many different positions the two 0's can take amongst the 1's. We know from the Binomial Theorem that the number of ways of choosing r objects from n objects is nCr = n!/r!(n-r)! In our case the 'objects' are the positions of the 1's and 0's and we want to know how many ways there are of choosing two positions from the fourteen available positions. Hence we can substitute into the above formula with n=14 and r=2 to find that the total number of solutions is 14C2=14!/(2!*12!)=91.