The positive integers $a, b, $ and $c$ are such that $ a + b + c =21$.
$a = 5, b = 5, c = 11$ is a solution. $a = 5, b = 11, c = 5$ is another solution (i.e. order matters).
How many different solutions are there?
I can see a very slow way of writing each individual case out, but this is definitely not ideal... Is there a faster way?
On
Here's a way to think about this:
Do you remember how tally's work: | = 1 || = 2 and ||| = 3.. and so forth. Let's take that but ||||| = 5, there are no diagonals.
There are going to be 21 tallies and 2 plus signs to achieve the sum of 21, so of 23 spots, we choose 21 to be the places we put the tallies. Or we can choose 2 of 23 spots to choose the plus signs.
But wait! The numbers - a, b, c - have to be positive as you specified. So let's declare variables $a = a_* + 1$, $b = b_* + 1$, $c = c_* + 1$.
Substituting this into the equation: $a + b + c = 21$ $$a_* + b_* + c_* = 18$$
So there are 20 spots to choose from (18 bars and 2 "+"'s). So the answer to your question is ${{20} \choose 2} = {{20} \choose 18}$
On
Read this article on combinatorial compositions to understand how the general problem is solved. The answer is $$\binom{21-1}{3-1}=\binom{20}{2}=\frac{20\cdot 19}{2}=190.$$
If $a=1$ there are 19 ways to choose $b$ and then only one way to choose $c$.
If $a=2$ there are 18 ways to choose $b$ and then only one way to choose $c$.
If $a=3$ there are 17 ways to choose $b$ and then only one way to choose $c$.
This pattern clearly continues, so there are $19 + 18 + 17 + … + 1$ ways altogether. This is simply an arithmetic series with first term 19, and common difference 1, so its sum is (20 x 19)/2. Hence your answer is 190.