Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
20 Nov 2009, 05:40
2
This post received KUDOS
Expert's post
10
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
75% (hard)
Question Stats:
65% (03:27) correct
35% (02:39) wrong based on 201 sessions
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.
Re: Cans in refrigerator. [#permalink]
20 Nov 2009, 07:29
3
This post received KUDOS
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? _________________
Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
31 Jan 2016, 23:52
2
This post received KUDOS
Expert's post
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
Bunuel, I have an objection about this question. Please help me about my concerns.
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).
Please help.
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!. _________________
Re: Cans in refrigerator. [#permalink]
20 Nov 2009, 07:45
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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.
Re: Cans in refrigerator. [#permalink]
20 Jul 2013, 14:12
1
This post received KUDOS
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 ? _________________
Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
Re: Cans in refrigerator. [#permalink]
20 Jul 2013, 14:27
1
This post received KUDOS
Expert's post
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 \(C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455\).
Re: Cans in refrigerator. [#permalink]
20 Nov 2009, 06:09
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, 06:41, edited 1 time in total.
Re: Cans in refrigerator. [#permalink]
20 Nov 2009, 07:14
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.
Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
14 Sep 2013, 11:13
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
Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
14 Sep 2013, 11:17
Expert's post
eladavid wrote:
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
Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
16 Sep 2013, 11:41
Expert's post
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).
Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
18 Sep 2013, 00:06
Expert's post
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.
Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
18 Sep 2013, 00:51
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 _________________
Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
30 Dec 2013, 07:12
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
Answer is D
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)
gmatclubot
Re: There are 12 cans in the refrigerator. 7 of them are red and
[#permalink]
30 Dec 2013, 07:12
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...