Number of ways to invest $\$20,000$ in units of $\$1000$ if not all the money need be spent

Question

Working through a combinatorics section currently and am working on this $2$-part problem. I have solved part $a$ quickly and will provide my work below but am having some trouble with part $b$ and seek guidance in how to proceed.

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Question Statement

An investor has $\$20,000$ to be invested amongst $4$ possible investments. Each investment must be a unit of $\$1,000$. If all the money needs to be invested then how many investment strategies are available? What if not all the money need be invested?

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Part $a$

This is a simple application of the formula for distributing $n$ identical objects into $k$ distinct boxes. The formula being used is $${n+k-1 \choose k-1}$$ Where we note that $n = 20$ since this is the number of available units of $\$1,000$ for investment and $k = 4$, the number of investments. Hence we find $${23 \choose 3} = \frac{23!}{3!17!} = 12,113,640$$

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Part $b$

For this part, my first instinct is to check each combination from $\$0$ invested all the way up to $\$20,000$ and then sum the results. Which would look like: $$\sum_{n=0}^{20} {n + 3 \choose 3} = \sum_{n=0}^{20} \frac{(n+3)!}{3!(n-3)!}$$ However, the first $3$ terms of this sum would lead to negative factorials, which haven't been covered (though I imagine $-3! = -3 \cdot -2 \cdot -1$) and the odd terms would then lead to negative values for the overall term, which doesn't make sense in the context of counting combinations.

Am I on the right track with my line of thinking? What adjustments should I try and think about in order to make the sum representative of the reality of the problem? Or am I way off and the solution is something else entirely? Thanks for any insight provided!

Answer 1

I have a work similar to your approach in part $a$ and same way as lulu commented. Assumed $20$ identical coins to be distributed among $4$ persons $A,B,C,D$. It is not necessary to spend all of the conins.

Assume $a$ is the number of coins to $A$ and similarly to other $(b,c,d)$

$a+b+c+d\leq 20$

Assume a variable $p\in [0,20]$. And the solutions to the above equation are similar to those,

$$a+b+c+d+p=20$$

Now you can apply your result from part $a$ which will give. $24 \choose 4$

Assuming investing no money is considered a case.

Answer 2

A: By the way, for part $1$, though you have written the formula correctly, your computation is way off !

$\binom{23}{3} = 1771$

B: Here I put a slightly different interpretation to
"What if not all the money need be invested" to mean that at least one unit is invested,

so adding the slack variable, the formula becomes $$\binom{n+k}{k} - 1 = \binom{24}{4} - 1 = 10625$$

[The $-1$ is for all the money going into the slack variable ]