Combinatorial Analysis (Investments)

[Win_odd Dhamnekar]

We have $20,000 that must be invested among 4 possible opportunities. Each investment must be integral in units of $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. How many different investment strategies are available if (a) an investment must be made in each opportunity? (b) investments must be made in at least 3 of the 4 opportunities?

My answers: a) We have to invest $9000 in the 4 opportunities considering the minimal investments of $2000, $2000, $3000, $4000 that are required to be made in each of the investment namely $\displaystyle x_1,x_2, x_3, x_4$ respectively.

So, $\displaystyle x_1 + x_2 + x_3 + x_4 = \$9000$ Here n=4, r=9. So, using the formula $\displaystyle \binom{n +r -1}{r - 1}= \binom{12}{8}=495$possible different investment strategies available if an investment must be made in each opportunity.

b) If an investments must be made in at least 3 of the 4 opportunities, then there are $\displaystyle 495 + \binom{15}{12}=950$ possible different investment strategies, in case we invest in the opportunities $\displaystyle x_1 + x_2 + x_3=\$13000$ or $\displaystyle 495 + \binom{14}{11}= 859,$ possible different investment strategies in case we invest in the opportunities $\displaystyle x_1 +x_2 + x_4= \$12000$or $\displaystyle 495 + \binom{13}{10}= 781$ possible different investment strategies in case, we invest in the opportunities $\displaystyle x_1 + x_3 + x_4 =\$ 11000$ or $\displaystyle x_2 + x_3 + x_4=\$11000$

But author gave the answers for a) 220 and for b) 572.

How is that? My answers are correct or author's answers are correct?

 

[BigBeachBanana]

a)
I would approach this a little bit different. Let $x_1,x_2,x_3,x_4$ be the investment made in the i-th opportunity and $m_1=2,000,m_2=2,000,m_3=3,000,m_4=4,000$ be the minimal investment needed to be made in the i-th opportunity.
So
$x_1+x_2+x_3+x_4=20,000$ and $\sum_{i=1}^{4}m_i=11,000$.
Now let $D_i=x_i-m_i$, the difference between the investment and the minimal amount.
$D_1 + D_2+D_3+D_4=20,000-\sum_{i=1}^{4}m_i=9,000$. It follows that we have $n=9,r=4$.
Therefore,
$${9+4-1 \choose 4-1}={12 \choose 3}={12 \choose 9}=220$$

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b) I'm gonna recycle the variables here.
$x_1+x_2+x_3=20,000$
$D_1 + D_2+D_3=20,000-\sum_{i=1}^{4}m_i+m_4=9,000+m_4$. It follows that $n=9+m_4,r=3$.
Thus,
$${9+m_4+3-1\choose 3-1}={11 +m_4 \choose 2}$$Lastly, considering the case from a), the total number of ways is:
$$220+\sum_{i=1}^{4}{11+m_i\choose 2}=220+ {11+2 \choose 2}+ { 11+2 \choose 2}+{ 11+3\choose 2}+ {11+4 \choose 2}= 572$$