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# In how many ways is it possible to put seven apples and

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Manager
Joined: 24 Jun 2003
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Location: Moscow
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In how many ways is it possible to put seven apples and [#permalink]  11 Aug 2003, 06:59
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In how many ways is it possible to put seven apples and three oranges in two paper bags so that both bags to contain at least one orange and identical number of fruits?
SVP
Joined: 03 Feb 2003
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10C5=252 total fives out of ten, but we count them twice, for one five defines the other
so, it is 126

wrong fives (no oranges, all apples) 7C5=21
126-21=105
Manager
Joined: 24 Jun 2003
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stolyar wrote:
10C5=252 total fives out of ten, but we count them twice, for one five defines the other
so, it is 126

wrong fives (no oranges, all apples) 7C5=21
126-21=105

Stolyar,

Here's my line of thinking:

The ways in which this arrangement can be made are

BAG-1 BAG-2

1 Orng+4 App 2 Orng+3 App
2 Orng+3 App 1 Orng+4 App

Therefore, there are 2 ways in my opinion. What's the flaw in this argument?

The key difference is that I'm assuming all oranges to be similar and all apples to be similar, therefore no choosing is involved
SVP
Joined: 03 Feb 2003
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1 Orng+4 App OR 2 Orng+3 App (2 combinations) would be OK if you had 5 fruits. But you have 10.

Consider
you have 1 white and 1 black balls, in how many combinations you can take a pair of black and white? The only combination--one black plus one white.

Now you have 1 white and 10 black balls; the question is the same--how many combinations. One again?
Manager
Joined: 24 Jun 2003
Posts: 94
Location: Moscow
Followers: 1

Kudos [?]: 1 [0], given: 0

Stolyar's approach is the best. The answer is 105
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