Simple Investment Problem

Baker · science forum beginner Joined: 19 Jun 2006 Posts: 2

I have 20 thousand dollars that must be invested among 4 possible
opportunities. Each investment must be made in integral units of 1000
dollars and there are minimal investments that need to be made if one is
is to invest in these opportunities. The minimal investments are
2, 2, 3, and 4 thousand dollars. How many different investment
strategies are there if:

a) an investment must be made in each opportunity
b) investments must be made in 3 of the 4 opportunities.

According to the book "A First Course in Probability" by "Sheldon Ross"
the correct answer to part a is "220". This is the anser I got as follows:

Letting x1, x2, x3, and x4 be the opportunities where x1 >= 2, x2 >= 2,
x3 >= 3 and x4 >= 4 and assuming ALL the 20 thousand must be invested
among the 4 opportunites we get:

x1 + x2 + x3 + x4 = 20

Now, letting y1 = (x1 + 1) y2 = (x2 - 1), y3 = (x3 - 2) and y4 = (x4 - 3)
we have:

y1 + y2 + y3 + y4 = 20 - 1 - 1 - 2 - 3 = 13.

where yi > 0 for all i = 1 ... 4, Letting n = 13 and r = 4 we have

C(n-1,r-1) = C(13-1, 4-1) = C(12,3) = 220 which is the answer in the book.

Now part B asks "how many investment strategies are there if investments
must be made in 3 of the 4 opportunities". The answer in the book is
"552". I have tried solving this part numerous different ways and come up
with every answer but 552. Can anyone shed any light on how the author
arrives at "552" for part B of this relatively simple problem ?? I can't
seem to figure it out.

Thanks

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