Combination Theory Problem

Question

I have the following problem:

Suppose that there exists 4 baskets of fruits, one with mangoes, another with guavas, another with papaya and the other with apples. Each basket contains at least 20 fruits and the fruits of a same basket are considered equal.

Of how many forms can we choose 10 fruits?
Of how many forms can we choose 10 fruits if we must have at least one of each type?

In trying to solve it, i am seeing there are a number of variables involved which confuses me. But I know that:

$$ 20: \text{Amount of each type of fruit} \\ a: \text{Number of mangoes chosen} \\ b: \text{Number of guavas chosen} \\ c: \text{Number of papayas chosen} \\ d: \text{Number of apples chosen} \\ $$

For the first part of the question, I have derived that:

$$ a + b + c + d = 10 $$

And here I remain stuck. Could someone explain to me the concept of the problem and how to proceed in solving the problem?

Stefan4024 · Accepted Answer · 2017-12-26T20:36:17.697

2

HINT: Use the Stars and Bars principle to get that the wanted number in part a) is $\binom{13}{3}$, while the wanted number in part b) is $\binom{9}{3}$

edited Dec 26 '17 at 20:36

answered Dec 26 '17 at 20:24

Stefan4024

35,843

Why 13 for part a)? – Omari Celestine Dec 26 '17 at 20:32
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10 stars and 3 bars ;) – Dec 26 '17 at 20:35
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We have $4$ terms, whose sum is $10$. By the Stars and Bars principle we have $\binom{10+4-1}{3}$ ways to do it. – Stefan4024 Dec 26 '17 at 20:36

Combination Theory Problem

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