Combination Problem Involving Integers 1-9 to Reach a Sum of 10

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I received a word problem that goes like this.

A local kindergarten is thinking of making posters that show all the different ways of adding two or more integers from 1 to 9 to get a sum of 10. If there is enough space on each poster for up to 50 possible solutions, how many posters will the school need to make?

(Note: sums that contain the same number but in a different order are considered to be different; for example, 1 + 9 and 9 + 1 are two different solutions.)

What is the answer to this problem, and more importantly, how do I solve it?

To be clear, a solution such as,"1+1+1+1+1+1+1+1+1+1+1," is a valid one.

Please make your explanation somewhat thorough and be available to explain further. Thanks!

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There are 2 best solutions below

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Since the summands must be at least $1$ and at most $9$, a particular solution to the question corresponds to the placement of one or more addition signs in the nine spaces between successive ones in a row of ten ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ For instance, $$1 + 1 1 + 1 1 1 + 1 1 1 1$$ corresponds to the solution $1 + 2 + 3 + 4$. The number of such solutions is equal to the number of non-empty subsets of a nine-element set, which is $2^9 - 1$.

To see this, observe that we have two choices for each of the nine spaces, to place an addition sign or not place one. That gives us $2^9$ solutions. However, we cannot leave all the spaces empty since that would correspond to using the summand $10$, which is not permitted. Therefore, the number of admissible solutions is $2^9 - 1$.

This sort of problem is called a composition.

I will leave it to you to determine how many posters will be needed.

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On

There are fourty two partitions of 10. You could simply list them all, then count how many distinct permuatations each has.

10 has $1$

9 + 1 has $2$

8 + 2 has $2$

8 + 1 + 1 has $3$

7 + 3 has $2$

7 + 2 + 1 has $6$

7 + 1 + 1 + 1 has $4$

et cetera

2 + 2 + 2 + 2 + 2 has $1$

2 + 2 + 2 + 2 + 1 + 1 has $\binom 64$

2 + 2 + 2 + 1 + 1 + 1 + 1 has $\binom 73$

2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 has $\binom 82$

2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 has $9$

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 has $1$

But if you think about it a bit more, there is a cleaver solution.