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John has 7 bananas and 3 kiwis. In how many ways can John [#permalink]
01 Feb 2005, 18:23

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John has 7 bananas and 3 kiwis. In how many ways can John divide the 10 fruit between two parcels, if there has to be an equal total number of fruit in either parcel and so that there is at least one kiwi in each parcel.

1. 1 k + 4 b or 2 k + 3 b
3C1 * 7C4 = 3 x 35 = 105 possibilities
2. 2 k + 3 b or 1 k + 4 b
3C2 * 7C3 = 3 x 35 = 105 possibilities
3. parcel 1 is equal to parcel 2 => only 105 possibilities at all

We need to select 5 from 10, with at least 1 kiwi.

1 kiwi C(3,1)C(7,4)=3*7*6*5/6=105

And that's it. No need to do two kiwis since it is symmetric.

Can you explain a little more?

According to your result, it only answers this question:

In how many ways can John divide the 10 fruit between two parcels, if there has to be an equal total number of fruit in either parcel and so that there is only one kiwi in one parcel and two in the other.

It's the same. One in this bag means two in the other bag. And two in this bag means one in the other bag. Say you only have three fruits. I treat the following two case as the same outcome:
A has f1 and B has f2, f3
A has f2, f3 and B has f1

We need to select 5 from 10, with at least 1 kiwi.

1 kiwi C(3,1)C(7,4)=3*7*6*5/6=105

And that's it. No need to do two kiwis since it is symmetric.

Here's what I don't get. I see two possiblities:
A. 4 bananas and 1 kiwi (7c4 * 3c1) 105 possibilities
B. 3 bananas and 2 kiwi (7c3 * 3c2) 105 possibilities

total # of poss = 210

what am I missing? I understand that the two parcels are symmetric, but you still have different fruits in case one vs. case two.

box1- 7C3 AND 3C2 OR 7C4 AND 3C1 = 105 + 105 = 210 box2- 7C4 AND 3C1 OR 7C3 AND 3C2 = 105 + 105 = 210

TOTAL=210+210 = 420 possibilities.

now where am i going wrong??????????? can anybody tell me plz

parcel 1: kbbbb (case1) OR kkbbb(case2) = 105 p (you misused OR as AND)
parcel 2 : kkbbb(case1) OR kbbbb(case2) = 105 p (you misused OR as AND)

it is a combination problem that means it does not matter which parcel has the combination; it only matters the different way of combinations; in your approach you count the combination that appears in parcel 1 as well as the combination in parcel 2, but you should count it only once. maybe it helps

box1- 7C3 AND 3C2 OR 7C4 AND 3C1 = 105 + 105 = 210 box2- 7C4 AND 3C1 OR 7C3 AND 3C2 = 105 + 105 = 210

TOTAL=210+210 = 420 possibilities.

now where am i going wrong??????????? can anybody tell me plz

parcel 1: kbbbb (case1) OR kkbbb(case2) = 105 p (you misused OR as AND) parcel 2 : kkbbb(case1) OR kbbbb(case2) = 105 p (you misused OR as AND)

it is a combination problem that means it does not matter which parcel has the combination; it only matters the different way of combinations; in your approach you count the combination that appears in parcel 1 as well as the combination in parcel 2, but you should count it only once. maybe it helps

ways to pick 5 fruits from 10 - ways to pick 5 friuts with all 3 kiwis - ways to pick 5 fruits with no kiwis
=10C5-8C3-7C5
=175

Note in both parcels there should be atleast 1 kiwi.Meaning you can not have 3 kiwis picked in the first parcel(second parcel will have zero kiwis), andyou can not have no kiwis at all.

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