Win_odd Dhamnekar
Junior Member
- Joined
- Aug 14, 2018
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- 207
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?
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?