We have$\backslash (20,000$that must be invested among $4$possible opportunities. Each investment must be integral in units$\backslash )1000,$and there are minimal investments that need to be made if one is to invest in these opportunities. The minimal investments are $\backslash (2000,\backslash )2000,\backslash (3000,$and$\backslash )4000$. How many different investment strategies are available if

$\left(a\right)$an investment must be made in each opportunity?

$\left(b\right)$investments must be made in at least $3$of the $4$opportunities?

We have $20,000$that must be invested among $4$possible opportunities. Each investment must be integral in units$1000,$and there are minimal investments that need to be made if one is to invest in these opportunities. The minimal investments are$2000,2000,3000,$and$4000$.

Given:

We have to distribute $20$units of money into 4opportunities.

The first opportunities have a minimum investment of $2,2,3$and$4$units of money. The number of positive integer solutions to the equation:

localid="1652770277430" $x_1+x_2+\dots +x_k=n\text{is}\left(\begin{array}{l}n-1\\ k-1\end{array}\right)$

And $\left(\begin{array}{c}n+k-1\\ n-1\end{array}\right)$non-negative integer solutions to the aforementioned equation.

Investing in every opportunity

Let $x_i$be the number of units of money invested in the $i$-th opportunity.

localid="1652770292816" $$\begin{gathered} y_1-x_1=2 \\ {y}_2-x=2 \\ {y}_3-x_3=3 \\ {y}_4-x_4=4 \end{gathered}$$

The equation becomes:

localid="1652770304173" $$\begin{gathered} y_1+2+y_2+2+y_3+3+y_4+4=20 \\ {y}_1+y_2+y_3+y_4=9 \end{gathered}$$

$y_i$are all nonnegative integers, so the number of possible solutions islocalid="1652770312160" $\left(\begin{array}{c}9+4-1\\ 4-1\end{array}\right)=220$.

Since$y_i$uniquely defines$x_i,$there are220 solutions to the starting equation.

If we invest in every opportunity there are $220$possible ways to do it.

Investing in at least three, If we invest in at least three opportunities. We can either invest in all four opportunities or in precisely three.

The number of possible investment strategies if we invest into opportunities$1,2,$and $3$is the number of solutions to:

localid="1652770329886" $$\begin{gathered} y_1+2+y_2+2+y_3+3=20 \\ {y}_1+y_2+y_3=13 \end{gathered}$$

That is.

localid="1652888164816" $\left(\begin{array}{c}13+3-1\\ 3-1\end{array}\right)=105$

The number of possible investment strategies if we invest into opportunities$1,2,$and$4$is the number of solutions to:

localid="1652770341795" $$\begin{gathered} y_1+2+y_2+2+y_4+4=20 \\ {y}_1+y_2+y_4=12 \end{gathered}$$

That is.

localid="1652770349378" $\left(\begin{array}{c}12+3-1\\ 3-1\end{array}\right)=91$

The number of possible investment strategies if we invest into opportunities$1,3,$and $4$is the number of solutions to:

localid="1652770360830" $$\begin{gathered} y_1+2+y_3+3+y_4+4=20 \\ {y}_1+y_3+y_4=11 \end{gathered}$$

That is.

localid="1652770370493" $\left(\begin{array}{c}11+3-1\\ 3-1\end{array}\right)=78$

The number of possible investment strategies if we invest into opportunities $2,3,$and $4$is the number of solutions to:

localid="1652770383823" $$\begin{gathered} y_2+2+y_3+3+y_4+4=20 \\ {y}_2+y_3+y_4=11 \end{gathered}$$

That is.

localid="1652770394413" $\left(\begin{array}{c}11+3-1\\ 3-1\end{array}\right)=78$

So the final number, if we invest in precisely three opportunities, is$105+91+78+78=352$. At least three opportunities is a union of precisely three and more than three.

The number of possible investing strategies of investing in at least three opportunities is$220+352=572$.

Consider the investment strategies as the solution to an equation.

There are $220$investment strategies if we invest in all four opportunities and $572$strategies if we were to invest in at least three of them.