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From 3 apples, 4 plums and 2 grapefruits how many selections

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Director
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From 3 apples, 4 plums and 2 grapefruits how many selections [#permalink] New post 06 Mar 2005, 16:12
From 3 apples, 4 plums and 2 grapefruits how many selections can be made taking atleast one of each kind?

Please let me know the concept behind this.

Thanks!!
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 [#permalink] New post 06 Mar 2005, 16:40
assumimg fruits r considered diff....

pick 1 apple = 3C1 = 3 ways
2 apples....= 3C2 = 3 ways
3 apples = 3C3 = 1 ways

total ways for apple = 3+3+1 = 7 ways


pick 1 plum = 4C1 = 4 ways
2 plum ....= 4C2 = 6 ways
3 plum = 4C3 = 4 ways
4 plum = 4C4 = 1 ways

total ways for plum= 4+6+4+1 = 15 ways

pick 1 grapefruit = 2C1 = 2 ways
2 grapefruit ....= 2C2 = 1 ways


total ways for grapefruit = 3 ways


Total ways = 7*15*3 = 315
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 [#permalink] New post 06 Mar 2005, 18:29
Baner that is one way of doing it but in the book I was doing from he gave a very simple step (which I didnt understand)

(2^3 -1)(2^4-1)(2^2-1)=315
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reason [#permalink] New post 07 Mar 2005, 06:02
Think of it this way:

How many ways you can choose from 3 apples?

2^3=8

atleast one apple has to be selected, which means (2^3-1). Similarly apply for other.

Thanks
Shishir
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 [#permalink] New post 07 Mar 2005, 07:49
The notion 2^n is for the Yes/No answer. For each of the three apples, there are two choices: You can choose it (Y) or not choose it (N). So the total outcome for three apples would be 2^3. Now, since you have to choose at least one apple, this means you have to take out the one outcome where no apple is chosen (NNN). Therefore you get 2^3-1. Do the same thing for other fruits.

When the number of fruits for each type is not too big, baner's approach may be more straight forward.
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 [#permalink] New post 07 Mar 2005, 09:06
How r we doing this without knowing the total number of selections to be made....im confused
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 [#permalink] New post 07 Mar 2005, 09:25
Rupstar wrote:
How r we doing this without knowing the total number of selections to be made....im confused


The question itself is to find the total number of selections.

One of the selection with minimum number of fruits could be to select one from each category. Thus minimum fruits that can be choosen are 3 and maximum fruits that can be chosen are 9. So the number of fruits that can be choosen start from 3 upto 9. For each number the question need to know, how many ways can this selection be made. Example, 3 fruits can only be choosen in one way i.e 1 apple, 1 plum and 1 grapefruit. Similarly 4 fruits can be choosen in many ways

2 apples, 1 plums and 1 grapefruits
1 apples, 2 plums and 1 grapefruits
1 apples, 1 plums and 2 grapefruits

And so on.....

Hope I have not confused you further... :-)

Ketan
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 [#permalink] New post 07 Mar 2005, 09:33
ok now it makes sense
thanks mate!
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 [#permalink] New post 07 Mar 2005, 10:34
HongHu:

Thanks !!! Thats it I could now get the concept.

Baner : your process of evaluation is right as always but I didnt know why this 2^n concept is being used.......

Bionomial Theorem says

nC0 + nC1 + nC2 +...nCn = 2^n

So the total # of combinations of 'n' things (taken in portions/all) = 2^n
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 [#permalink] New post 08 Mar 2005, 07:49
HongHu wrote:
The notion 2^n is for the Yes/No answer. For each of the three apples, there are two choices: You can choose it (Y) or not choose it (N). So the total outcome for three apples would be 2^3. Now, since you have to choose at least one apple, this means you have to take out the one outcome where no apple is chosen (NNN). Therefore you get 2^3-1. Do the same thing for other fruits.

When the number of fruits for each type is not too big, baner's approach may be more straight forward.

this makes my concept clear..thanks honghu.
  [#permalink] 08 Mar 2005, 07:49
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