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31 Aug 2010, 11:10
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There are 4 different fruits in a basket with at least 2 of each type. What is the minimum number of combinations possible such that 3 fruits are picked up in a manner that 2 are of the same type and one of another type?
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31 Aug 2010, 19:03
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My attempt:
Since the basket contains at least 2 fruits of each variety and we need to determine the minimum combination, let us assume that there are 2 number of each variety.
Hence there are 4 fruits within the basket. A- 2, B- 2, C- 2 and D- 2
Assume the 2 fruit of the same variety (A). Hence combination is
2C1 * 1 * 6C1 (One out of the remaining 3 variety i.e., One out of 3*2 fruits)
12.
Since the basket contains at least 2 fruits of each variety and we need to determine the minimum combination, let us assume that there are 2 number of each variety.
Hence there are 4 fruits within the basket. A- 2, B- 2, C- 2 and D- 2
Assume the 2 fruit of the same variety (A). Hence combination is
2C1 * 1 * 6C1 (One out of the remaining 3 variety i.e., One out of 3*2 fruits)
12.
31 Aug 2010, 20:27
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ezhilkumarank
My solution was 8 X 1 X 6 = 48 (because there are a min of 8 fruits and any one can be the first fruit to be drawn. 1 because the next fruit should be the same type and there is a minimum of one such fruit that can take the seconf place and 6 of the remaining fruits can be picked for the third place)
I dont know what is the correct answer. Also would the answer be any different if the minimum combination is not mentioned and we need to simply find the number of combinations possible? Because in the original question I dont remember seeing the word minimum. I just put it there since it made sense.
