Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 39672

There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
20 Nov 2009, 06:40
4
This post received KUDOS
Expert's post
10
This post was BOOKMARKED
Question Stats:
66% (03:20) correct
34% (02:44) wrong based on 291 sessions
HideShow timer Statistics



Manager
Joined: 13 Oct 2009
Posts: 118
Location: USA
Schools: IU KSB

Re: Cans in refrigerator. [#permalink]
Show Tags
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.



Director
Joined: 01 Apr 2008
Posts: 881
Name: Ronak Amin
Schools: IIM Lucknow (IPMX)  Class of 2014

Re: Cans in refrigerator. [#permalink]
Show Tags
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.



Director
Joined: 01 Apr 2008
Posts: 881
Name: Ronak Amin
Schools: IIM Lucknow (IPMX)  Class of 2014

Re: Cans in refrigerator. [#permalink]
Show Tags
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.



Senior Manager
Joined: 21 Jul 2009
Posts: 364
Schools: LBS, INSEAD, IMD, ISB  Anything with just 1 yr program.

Re: Cans in refrigerator. [#permalink]
Show Tags
20 Nov 2009, 08:29
4
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?
_________________
I am AWESOME and it's gonna be LEGENDARY!!!



Math Expert
Joined: 02 Sep 2009
Posts: 39672

Re: Cans in refrigerator. [#permalink]
Show Tags
20 Nov 2009, 08:45
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED



Intern
Joined: 20 Nov 2009
Posts: 14

Re: Cans in refrigerator. [#permalink]
Show Tags
20 Nov 2009, 11:02
its 7C6*5C2+7C5*5C3+7C4*5C4=70+210+175=455



Manager
Joined: 11 Sep 2009
Posts: 100

Re: Cans in refrigerator. [#permalink]
Show Tags
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



Current Student
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 961
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

Re: Cans in refrigerator. [#permalink]
Show Tags
20 Jul 2013, 15:12
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".



Math Expert
Joined: 02 Sep 2009
Posts: 39672

Re: Cans in refrigerator. [#permalink]
Show Tags
20 Jul 2013, 15:27
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 \(C^8_{12}(C^7_7*C^1_5+C^5_5*C^3_7)=495(5+35)=455\). Answer: D.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 05 Feb 2013
Posts: 13

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



Math Expert
Joined: 02 Sep 2009
Posts: 39672

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
14 Sep 2013, 12:17



Intern
Joined: 10 Aug 2013
Posts: 20

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
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".
_________________
The First and Last time !!! BKPL  Below Kudos Poverty Line .....Need your help.



Math Expert
Joined: 02 Sep 2009
Posts: 39672

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 10 Aug 2013
Posts: 20

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
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
_________________
The First and Last time !!! BKPL  Below Kudos Poverty Line .....Need your help.



Math Expert
Joined: 02 Sep 2009
Posts: 39672

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
18 Sep 2013, 01:06



Director
Joined: 17 Dec 2012
Posts: 549
Location: India

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
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
_________________
Srinivasan Vaidyaraman Sravna http://www.sravnatestprep.com
Classroom and Online Coaching



Manager
Status: Persevering
Joined: 15 May 2013
Posts: 218
Location: India
Concentration: Technology, Leadership
GMAT Date: 08022013
GPA: 3.7
WE: Consulting (Consulting)

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
22 Sep 2013, 07:14
Yep, did it the other way around 7c6*5c2+7c5*5c3+7c4*5c4=455 Hence,D
_________________
It's one thing to get defeated, but another to accept it.



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

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



Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance

Re: There are 12 cans in the refrigerator. 7 of them are red and [#permalink]
Show Tags
24 Apr 2014, 15:39
Here's another way. Although I think this will require some sanity check by Bunuel Anways Total number of ways to pick 8 from 12 is: 12C8 = 495 Now then, we are working with restrictions then: Total number of ways NOT at least 1 red is 5 Total number of ways NOT at least 1 blue is 35 Total restrictions is 40 Therefore 49540=455 Answer is thus D Is this method valid? Cheers! J




Re: There are 12 cans in the refrigerator. 7 of them are red and
[#permalink]
24 Apr 2014, 15:39



Go to page
1 2
Next
[ 29 posts ]





Similar topics 
Author 
Replies 
Last post 
Similar Topics:


10


Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an

Bunuel 
8 
18 May 2017, 20:10 

1


In a group of 20 people, 5 of them belong to the golf club, 7 to the

Bunuel 
4 
06 May 2017, 17:33 

2


In How many ways can 7 people park cars in slots allotted to them?

susheelh 
1 
02 Jul 2016, 04:51 

1


In the "BigReds" parking lot there are 56 vehicles, 18 of them are

Bunuel 
2 
20 Jun 2016, 01:00 

11


There are 12 red balls, 10 blue balls, 15 red balls and 9

cyberjadugar 
7 
15 Jun 2012, 02:43 



