How many ways are there to have a collection of eight fruits from a large pile of identical oranges, apples, bananas, peaches, and pears if the collection should include exactly two different kids of fruits?

I came up with 90 ways. Is this correct? If not, could have only hints on how to get the correct answer?


The answer is 70:

1) If there are 5 kinds of fruit, there are 10 different pairs of fruit kinds (4+3+2+1). The questions specifies that you need to select exactly two kinds of fruit to build the collection.

2) From 8 fruit there are 7 ways to choose how many of the 1st kind to choose and how many from the 2nd kind to choose (7+1, 6+2, 5+3, 4+4, 3+5, 2+6, 1+7). So you'd choose 7 apples and 1 orange, for example.

Note: These are not 9 ways as I suppose you calculated, because if you choose 8+0 or 0+8 of each fruit, you'll have only 1 kind, not as the question says exactly two kinds.

multiply the probabilities 7*10 and you get 70 :)

  • $\begingroup$ Yeah, I did calculate the cases where one would have none selected. Thanks. $\endgroup$ – gandolf Oct 4 '13 at 8:51

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