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John has 7 bananas and 3 kiwis .he needs to divide the 10 in

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Manager
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John has 7 bananas and 3 kiwis .he needs to divide the 10 in [#permalink] New post 30 Nov 2004, 19:02
John has 7 bananas and 3 kiwis .he needs to divide the 10 in 2 parcels
so that there is an equal total number of fruit in either parcel and so that there is at least one kiwi in each parcel.
How many ways ?

a) 21
b) 35
c) 60
d) 105
e) 120

Please explain the answer
Director
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 [#permalink] New post 30 Nov 2004, 20:54
Should the last choice read 210 instead of 120. Pls confirm.

Anyways, I think..

a) There has to be 5 in each parcel with atleast 1 kiwi in each of them.
b) In such case, it would be to choose the first 5 in the first parcel and the parcel 2 will only have what is left automatically.
c) So, ways to choose the first parcel would be

3C1*7C4 + 3C2*7C3 = 210.

3C1 (one kiwi out of 3) * 7C4 (because remaning 4 fruits from the 7 bannas) OR 2 kiwis and 3 bananas. There cant be a 3 kiwi situation as the restraint is that there got to be atleast one kiwi in a parcel.

Am I missing anything here..?
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 [#permalink] New post 30 Nov 2004, 21:30
Should the last choice read 210 instead of 120. Pls confirm.
-- No , it is 120 as given.

Nothing is missing in the question..........
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 [#permalink] New post 30 Nov 2004, 22:05
My bad,

If we choose 3C1*7C4 in one of the parcels, the other parcel will naturally have what is left and since it does not matter which parcel has the combination above, it will simply be 105.
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 [#permalink] New post 30 Nov 2004, 22:44
OA is D) ..You are right 105.

Your explanation is easier to understand as compared to the explanation in original source....

But don't we take into account the combinations for other parcel...pls. elaborate little more...

Thanks !
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 [#permalink] New post 30 Nov 2004, 22:46
venksune wrote:
My bad,

If we choose 3C1*7C4 in one of the parcels, the other parcel will naturally have what is left and since it does not matter which parcel has the combination above, it will simply be 105.


Not 100% on this one, but I think I agree with venksune.

Notice that if you work out the combination in the other direction, meaning, if you work out how many ways to have 2 kiwi and 3 bananas, it also comes out to 105.

where is this question from?
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 [#permalink] New post 01 Dec 2004, 00:02
got it from gmat club...bought some questions from gmat club.....
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 [#permalink] New post 01 Dec 2004, 06:45
it would be either 3c2 * 7c3 ways or 3c1 * 7c4 ways which is the same.
105.
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 [#permalink] New post 01 Dec 2004, 07:44
My first answer was 1... as there is no clue that these fruits are dinstinct...

Since we having only 2 parcels, allocating fruits to the first one will lead to the same number of ways as allocating fruits to the second one.

1st one (2K, 3B) = 3C2.7C3 = 3.35 = 105
or
2nd one (1K, 4B) = 3C1.7C4 = 3.35 = 105
  [#permalink] 01 Dec 2004, 07:44
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