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gmat2me2
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mckenna
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3 apples 4 plums 2 grapefruit.

1 grapefruit, x plums, y apples = X:
2C1.3C1(4C1 + 4C2+4C3+4C4) + 2C1.3C2(4C1+4C2+4C3+4C4) + 2C1.3C3(4C1+4C2+4C3+4C4)


2 grapefruits, x plums, y apples = Y :?
2C2.3C1(4C1 + 4C2+4C3+4C4) + 2C2.3C2(4C1+4C2+4C3+4C4) + 2C2.3C3(4C1+4C2+4C3+4C4)

Total = X + Y.

Seemingly simple problem, but mushroomed into a bunch of calculations.
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gmat2me2
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I think there is a very simple solution to this problem. Just waiting for the other guys to respond
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mckenna
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From 3 apples, 4 plums and 2 grapefruits how many selections can be made taking atleast one of each kind?

If all fruits of a kind are assumed to be identical, the answer is 3x4x2 = 24

If not, then the answer is (2^3 - 1) x (2^4 - 1) x (2^2 - 1) = 7x15x3 = 315
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The problem is to select at least one of the all items. So the actual number of choces are reduced by one.

No of choices will be = (No of Apples - 1) * (No of Plums -1) * (No of Grape fruit -1)



i.e. = 2 * 3 * 1 = 6

Please let me know whether this is correct.

regards,
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mckenna
From 3 apples, 4 plums and 2 grapefruits how many selections can be made taking atleast one of each kind?

If all fruits of a kind are assumed to be identical, the answer is 3x4x2 = 24

If not, then the answer is (2^3 - 1) x (2^4 - 1) x (2^2 - 1) = 7x15x3 = 315


Great post mckenna!

For previous discussions see:
https://www.gmatclub.com/phpbb/viewtopic.php?p=89346



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