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John has 7 bananas and 3 kiwis. In how many ways can John

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John has 7 bananas and 3 kiwis. In how many ways can John [#permalink] New post 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.

Pls. show your work.

Thanks.
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 [#permalink] New post 01 Feb 2005, 19:58
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.
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 [#permalink] New post 01 Feb 2005, 23:58
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
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 [#permalink] New post 02 Feb 2005, 00:19
HongHu wrote:
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.
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 [#permalink] New post 02 Feb 2005, 00:22
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

Will have to confirm with the OA though.
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 [#permalink] New post 02 Feb 2005, 00:52
But that only applies when the bags are identical though. Right?
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 [#permalink] New post 02 Feb 2005, 07:43
qhoc0010 wrote:
But that only applies when the bags are identical though. Right?


Yes, applies to this case since we are making two parcels, nothing special with either of them.
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594? [#permalink] New post 04 Feb 2005, 10:34
7B + 3K in two parcels say I and II

No of ways :
(a) 1K in I and 1K in II == 3*2 = 6 ways

(b) 1K1B in I and 1K1B or 2K in II and vice versa = = 3*7*(2*6+2)*2=21(14)2=588 ways

Add 594 ways
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Clarification [#permalink] New post 04 Feb 2005, 10:38
I assumed that the bananas and kiwis are not identical i mean in terms of shape size or color. So 7 distinct bananas and 3 distinct Kiwis.
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 [#permalink] New post 17 Feb 2005, 19:20
HongHu wrote:
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.
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 [#permalink] New post 18 Feb 2005, 20:41
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
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 [#permalink] New post 19 Feb 2005, 10:22
faizaniftikhar83 wrote:
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 :-D
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silly question [#permalink] New post 19 Feb 2005, 14:31
What if it had been without the limitation of at least 1 kiwi per parcel?
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 [#permalink] New post 21 Feb 2005, 05:59
christoph wrote:
faizaniftikhar83 wrote:
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 :-D


still...we get 210 ways....then howcome 105????
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 [#permalink] New post 21 Feb 2005, 07:15
hmmm. earlier I calculated 210 but then I saw explainations and understood why it is 105.
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 [#permalink] New post 21 Feb 2005, 14:08
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.
  [#permalink] 21 Feb 2005, 14:08
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