Question

An investor has 15 thousand dollars to invest among 3 possible investments A, B, C. Each investment must be in units of a thousand dollars. Not all the money need be invested. He must invest at least 3 thousand dollars on the investment A and at least 5 thounsand dollars on the investment B. How many different investment strategies are possible

Answer 1

Answer:

36 ways

Explanation:

There are 3 possible investments A, B and C.

If x represents the amount of money to be invested in the ith opportunity, it is given as:

$x_1+x_2+x_3=15$

Let y be the minimum investment to be made, therefore:

$y_1=x_1-3\\y_2=x_2-5\\y_3=x_3\\\\Therefore:\\\\y_1+y_2+y_3+3+5=15\\\\y_1+y_2+y_3+8=15\\\\y_1+y_2+y_3=7$

The number of possible combinations is:

C(n + k - 1, k - 1)

Where k is the number of investment = 3 and n = 7. Therefore:

The number of possible combinations is = C(n + k - 1, n - 1) = C(7 + 3 -1, 3 - 1) = C(9, 2) = $\frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\ ways$