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# There are 12 cans in the refrigerator. 7 of them are red and

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There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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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
[Reveal] Spoiler: OA

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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: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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01 Feb 2016, 00:52
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leve wrote:
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

If we are asked the number of ways removing the cans from a refrigerator then physically all cans must be taken indivudually and considered distinct! Although they look the same all cans has a distinct position in the refrigerator and taking out the cans is completly a physical thing. So order should matter in the whole process -from beginnig to the end! (Other wise the question should state that. But it doesnt.)

So we have 12 distinguisable cans (by their positions) which are R1, R2, R3, R4, R5, R6, R7 and B1, B2, B3, B4, B5 (Because they are physical and all has a distinct position in the refrigerator)

Solution from this point of view is;

There are 3 cases: (6R, 2B), (5R, 3B), (4R, 4B)

(6R, 2B): 7C6 * 5C2 * 8! (7C6 is number of ways choosing 6 red cans out of 7, 5C2 is number of ways choosing 2 blue cans out of 5 and 8! is the total number of orders/permutations.
(6R, 2B): 7C5 * 5C3 * 8!
(4R, 4B): 7C4 * 5C4 * 8!

So i believe the answer should be (70+210+175) * 8! = 455 * 8! not 455.

Moreover I believe 455 by it self is a meaningless answer! Because in this approach the cans are taken distinct in their sub sets (red group and blue group) but not considered distinct as a whole (red and blue groups mixed).

Responding to a pm:

All the cans are considered distinct.

Forget this question for a minute - consider this one:
There are 12 students in a class. In how many ways can you select 8 of them for a group activity?
How do you calculate in this case? You use 12C8 and that is all, right? Why? Do you say that we must multiply it by 8! because the students are sitting in a class in a particular way? No. We have already considered that the students are distinct. That is why from 12 distinct students we select 8. Now their arrangement in the class doesn't affect anything. Only if after selecting 8 students, we need to arrange them in 8 different spots, then we multiply by 8!.

Similarly, all cans are considered distinct. Say they are numbered R1, R2, R3... B1, B2... Now all you have to do is select 8 out of these 12. Their actual arrangement in the fridge is immaterial. Also, the question doesn't say that we need to arrange the 8 cans in 8 spots after removing them from the fridge. So we will not multiply by 8!.
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20 Nov 2009, 08:45
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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.

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20 Jul 2013, 15:12
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I considered all cans are identical in respective color group, ensuring at least 1R and 1B by considering these two groups ( 6R, 4B )

Case 1: 6R 2B ( blue will naturally take empty space ) thus we should calculate only 8C6 to mark possible slots on 8 available turns i.e = 28
Case 2: 5R 3B similarly 8C5 = 56.
Case 3: 4R 4B similarly 8C4 = 70.
<I can not further decrease R below 4 as it will take 3R 5B which violates the criterion>

E.g R _ R _ R _ R _ or RR_ _ R_ _ or _ _ R_R_RR (we can see empty slots are available for blue by default)
thus combination for Red cans on 8 slots is sufficient, as per my understanding.

But this approach gives me result as 28 + 56 + 70 = 154.
Which is not there in the options, is my approach right ?
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20 Jul 2013, 15:27
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PiyushK wrote:
I considered all cans are identical in respective color group, ensuring at least 1R and 1B by considering these two groups ( 6R, 4B )

Case 1: 6R 2B ( blue will naturally take empty space ) thus we should calculate only 8C6 to mark possible slots on 8 available turns i.e = 28
Case 2: 5R 3B similarly 8C5 = 56.
Case 3: 4R 4B similarly 8C4 = 70.
<I can not further decrease R below 4 as it will take 3R 5B which violates the criterion>

E.g R _ R _ R _ R _ or RR_ _ R_ _ or _ _ R_R_RR (we can see empty slots are available for blue by default)
thus combination for Red cans on 8 slots is sufficient, as per my understanding.

But this approach gives me result as 28 + 56 + 70 = 154.
Which is not there in the options, is my approach right ?

Obviously not.

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

Total ways to select 8 cans out of 7+5=12 is $$C^8_{12}$$;
Ways to select 8 cans so that zero red cans are left is $$C^7_7*C^1_5$$;
Ways to select 8 cans so that zero blue cans are left is $$C^5_5*C^3_7$$;

Hence ways to select 8 cans so that at least one red can and at least one blue can to remain is $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

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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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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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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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20 Nov 2009, 11:02
its 7C6*5C2+7C5*5C3+7C4*5C4=70+210+175=455

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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

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There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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14 Sep 2013, 12:13
I don't understand how did you guys get to the numerical numbers

[wrapimg=][/wrapimg]

12C4 - 7C4 - 5C4 = 495 - 35 -5 = 455[wrapimg=][/wrapimg]

Last edited by eladavid on 02 Apr 2016, 04:49, edited 3 times in total.

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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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14 Sep 2013, 12:17
I don't understand how did you guys get to the numerical numbers
I mean I know it might be a basic question but i don't see the connection between the values

[wrapimg=][/wrapimg]

12C4 - 7C4 - 5C4 = 495 - 35 -5 = 455[wrapimg=][/wrapimg]

$$C^n_k=\frac{n!}{k!(n-k)!}$$

Check here: math-combinatorics-87345.html
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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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16 Sep 2013, 12:10
Bunnel I don't get it. Question says ...that at least one red and at least one blue cans to remain the refrigerator...

I think D would be answer if question were ...that at least one red OR at least one blue cans to remain the refrigerator...notice OR.

please enlighten me if I am going to wrong way.

Also what will be the answer if it were "OR" instead of "And".
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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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16 Sep 2013, 12:41
StormedBrain wrote:
Bunnel I don't get it. Question says ...that at least one red and at least one blue cans to remain the refrigerator...

I think D would be answer if question were ...that at least one red OR at least one blue cans to remain the refrigerator...notice OR.

please enlighten me if I am going to wrong way.

Posted from my mobile device

The question asks: 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.

So, we can remove:
6 red, 2 blue (1 red, 3 blue are left) --> $$C^2_5*C^6_7=70$$;
5 red, 3 blue (2 red, 3 blue are left) --> $$C^3_5*C^5_7=210$$;
4 red, 4 blue (3 red, 1 blue are left) --> $$C^4_5*C^4_7=175$$;

70+210+175=455.

The way it's olved in my post is different: (total)-(restriction).

Hope it helps.
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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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16 Sep 2013, 12:51
I got it...Thanks.

and one thing, ways in which at least one blue OR at least one red remains would be 12c8 . Isn't it ?

Posted from my mobile device
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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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18 Sep 2013, 01:06
StormedBrain wrote:
I got it...Thanks.

and one thing, ways in which at least one blue OR at least one red remains would be 12c8 . Isn't it ?

Posted from my mobile device

Yes. In this case no matter which 8 cans we remove from 12, there still will be at least 1 red or at least one blue can remaining in the ref: 12C8=495.

In this case we could also go the long way: 3 cases from my previous post + 2 more cases:
7 red, 1 blue (0 red, 4 blue are left) --> $$C^1_5*C^7_7=5$$;
3 red, 5 blue (4 red, 0 blue are left) --> $$C^5_5*C^3_7=35$$;

Total = 455(from my previous post) + 35 + 5 = 495.

Hope it's clear.
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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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18 Sep 2013, 01:51
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

1. Selection has to be made from each of two distinct groups, being red cans and blue cans

2. n1=7, n2=5

3. r1= 6 and r2=2 or
r1=5 and r2=3 or
r1=4 and r3=4,

each of which satisfies that there is at least one red can and one blue can left.

The total number of ways = 7C6*5C2 + 7C5*5C3 + 7C4*5C4 = 455
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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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22 Sep 2013, 07:14
Yep, did it the other way around

7c6*5c2+7c5*5c3+7c4*5c4=455 Hence,D
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Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]

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30 Dec 2013, 08:12
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

7C6*5C3 + 7C4*5C3 + 7C4*5C4 = 455

Hope it helps!
Cheers!

PS. I suggest listing range of values of each can just to visualize problem better
Then just pick the values that add up to 8
Namely, (2+6), (3+5) and (4+4)

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Re: There are 12 cans in the refrigerator. 7 of them are red and   [#permalink] 30 Dec 2013, 08:12

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