# In how many ways can we list out a string of seven numbers (using only the numbers 1, 2, and 3) such that they sum to 10?

77

#### Explanation:

Let's first look at the number of ways we can have 7 digits with the number 1, 2, and 3 add up to 10:

$3 + 2 + 1 + 1 + 1 + 1 + 1 = 10$
$2 + 2 + 2 + 1 + 1 + 1 + 1 = 10$

In fact, there are no other combinations of numbers that will get us to 10.

Ok, so how many permutations can we make with these combinations of numbers?

Let's do $3 + 2 + 1 + 1 + 1 + 1 + 1 = 10$ first:

There are 7 places the number 3 can go - so that is 7.

Once the 3 is placed, there are 6 places the 2 can go - so that is 6.

The rest of the numbers are 1's and so there is only 1 way to do the "filler" with the 1s.

So there are $7 \times 6 = 42$ different numbers we can make with this combination.

Now to $2 + 2 + 2 + 1 + 1 + 1 + 1 = 10$

We can place the three 2s across the seven places in many different ways. This is a combinations problem (we don't care which 2 ends up where), and so there are:

C_(7,3)=(7!)/((3!)(7-3)!)=(7!)/((3!)(4!) ways, which equals:

(7xxcancel6xx5xxcancel(4!))/(cancel(3xx2)xxcancel(4!))=35

And so there are $42 + 35 = 77$ different numbers that can be made.