How Many Investment Strategies Exist for $20,000 Across 4 Opportunities?

[e(ho0n3]

[SOLVED] Funky Counting Question

Problem. We have 20 thousand dollars that must be invested among 4 possible opportunities. Each investment must be integral in units of 1 thousand dollars, and there are minimal investments that need to be made if one is to invest in these opportunities. The minimal investments are 2, 2, 3, and 4 thousand dollars. How many different investment strategies are available if investments must be made in at least 3 of the 4 opportunities.

Answers. 572

Attempt. Let a, b, c and d be the 4 possible opportunities with minimal investments of 2, 2, 3 and 4 thousand dollars respectively. The possible combinations of opportunities are abc, abd, acd and abcd right? Now, would it illegal to invest 3 thousand dollars into opportunity a? Must it be multiples of 2 thousand? Ditto for the rest of the opportunities.

If it has to be in multiples of the minimum investment, how would one count them effectively? All I can think of doing is writing down combinations and looking for patterns.

 

[learningphysics]

Hint: For each case... subtract away the minimum investment and see the numbers of ways to distribute the rest.

 

[e(ho0n3]

OK. I realize my mistake now. Also, there are more than 4 possible combinations of opportunities. The combinations are abc, abd, acd, bcd, abcd.

Consider abc. After investing the minimum, I'm left with 13 thousand to invest. Let A, B, C be the quantities in thousands that I must invest after the minimum investment in a, b, c respectively. All I have to do is find the number of nonnegative integer solutions to A + B + C = 13 right? That is C(13 + 3 - 1, 3 - 1) = C(15, 2) = 105. For abd, I get C(12 + 3 - 1, 3 - 1) = C(14, 2) = 91. For acd, I get C(11 + 3 - 1, 3 - 1) = C(13, 2) = 78. bcd is the same as acd so it should be 78. For abcd I get C(9 + 4 - 1, 4 - 1) = C(11, 3) = 165. I get 517 after adding them all up.

I'm still missing something. What could I be missing?

 

[learningphysics]

For abcd I get C(9 + 4 - 1, 4 - 1) = C(11, 3) = 165.

This should be (12,3) = 220

 

[e(ho0n3]

Ah! How embarrassing. Thanks.

 

[learningphysics]

Ah! How embarrassing. Thanks.

no prob.