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# Cans in refrigerator.

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Cans in refrigerator. [#permalink]  20 Nov 2009, 06:40
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There are 12 cans in the refrigerator. 7 of them are red and 5 of them are blue. In how many ways we can remove 8 cans so that at least one red and at least one blue cans to remain the refrigerator.

(A) 460
(B) 490
(C) 493
(D) 455
(E) 445
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Re: Cans in refrigerator. [#permalink]  20 Nov 2009, 07:09
Bunuel wrote:
There are 12 cans in the refrigerator. 7 of them are red and 5 of them are blue. In how many ways we can remove 8 cans so that at least one red and at least one blue cans to remain the refrigerator.

(A) 460
(B) 490
(C) 493
(D) 455
(E) 445

'D' - 455

Ways to pick 4 cans so that at least one red and at least one blue cans to remain the refrigerator =
total ways to pick 4 can out of 12 - ways to pick 4 red out of 7 red - ways to pick 4 blue out of 5 blue

12C4 - 7C4 - 5C4 = 495 - 35 -5 = 455

Last edited by swatirpr on 20 Nov 2009, 07:41, edited 1 time in total.
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Re: Cans in refrigerator. [#permalink]  20 Nov 2009, 07:14
C for me.

Basically the question is asking us to find the number of ways in which 4 cans can be selected such that atleast one is red and atleast one is blue.

12C4 = 495. From this we will subtract 2 possibilities: i.e. all are blue and all are red. So we get 493.
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Re: Cans in refrigerator. [#permalink]  20 Nov 2009, 08:14
swatirpr wrote:
'D' - 455

Ways to pick 4 cans so that at least one red and at least one blue cans to remain the refrigerator =
total ways to pick 4 can out of 12 - ways to pick 4 red out of 7 red - ways to pick 4 blue out of 5 blue

12C4 - 7C4 - 5C4 = 495 - 35 -5 = 455

Sounds logical. Question: Do we know that we always have 7 red cans to choose from?? Out of these 7..4 can be in the 'chosen 8 cans'...so we are left with only 3 red cans to choose from.
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Re: Cans in refrigerator. [#permalink]  20 Nov 2009, 08:29
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Logic wise does this logic work?

Total number of ways to selecting 8 cans from 12 cans MINUS [ number of ways of leaving no red can at all in choosing 8 cans PLUS number of ways of leaving no blue can at all in choosing 8 cans]

12C8 - [ (7C7*5C1) + (5C5*7C3)] = 495 - 40 = 455.

D is the answer, what is the OA?
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Re: Cans in refrigerator. [#permalink]  20 Nov 2009, 08:45
SensibleGuy wrote:
Logicwise does this logic work?

Total number of ways to selecting 8 cans from 12 cans MINUS [ number of ways of leaving no red can at all in choosing 8 cans PLUS number of ways of leaving no blue can at all in choosing 8 cans]

12C8 - [ (7C7*5C1) + (5C5*7C3)] = 495 - 40 = 455.

D is the answer, what is the OA?

No need to postpone the correct answer. It's my question, so only my solution is available. And it's exactly the one Barney proposed. Perfect logic +1.

Answer: D.
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Re: Cans in refrigerator. [#permalink]  20 Nov 2009, 11:02
its 7C6*5C2+7C5*5C3+7C4*5C4=70+210+175=455
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Re: Cans in refrigerator. [#permalink]  13 Dec 2009, 03:31
My take is D

Any combination of 4 cans – combination without red cans – combination without blue cans = 12C4 – 7C4 – 5C4 = 455
Re: Cans in refrigerator.   [#permalink] 13 Dec 2009, 03:31
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