GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 30 Mar 2020, 11:06 ### GMAT Club Daily Prep

#### 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

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

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 62352
There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

6
11 00:00

Difficulty:   65% (hard)

Question Stats: 68% (02:54) correct 32% (02:50) wrong based on 212 sessions

### HideShow timer Statistics

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

_________________
Senior Manager  Joined: 21 Jul 2009
Posts: 267
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

### Show Tags

7
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?
##### General Discussion
Manager  Joined: 13 Oct 2009
Posts: 82
Location: USA
Schools: IU KSB

### Show Tags

1
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

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

Originally posted by swatirpr on 20 Nov 2009, 06:09.
Last edited by swatirpr on 20 Nov 2009, 06:41, edited 1 time in total.
Director  Joined: 01 Apr 2008
Posts: 584
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

### Show Tags

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: 584
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 62352

### Show Tags

1
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.

_________________
Intern  Joined: 19 Nov 2009
Posts: 4

### Show Tags

its 7C6*5C2+7C5*5C3+7C4*5C4=70+210+175=455
Intern  Joined: 11 Sep 2009
Posts: 47

### Show Tags

My take is D

Any combination of 4 cans – combination without red cans – combination without blue cans = 12C4 – 7C4 – 5C4 = 455
Director  Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 689
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

### Show Tags

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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 62352

### Show Tags

1
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 $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

_________________
Intern  Joined: 05 Feb 2013
Posts: 13
There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

I don't understand how did you guys get to the numerical numbers

[wrapimg=][/wrapimg]

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

Originally posted by eladavid on 14 Sep 2013, 11:13.
Last edited by eladavid on 02 Apr 2016, 03:49, edited 3 times in total.
Math Expert V
Joined: 02 Sep 2009
Posts: 62352
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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
_________________
Intern  Joined: 10 Aug 2013
Posts: 14
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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".
Math Expert V
Joined: 02 Sep 2009
Posts: 62352
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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.
_________________
Intern  Joined: 10 Aug 2013
Posts: 14
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 62352
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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.
_________________
Director  S
Joined: 17 Dec 2012
Posts: 619
Location: India
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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
Magical Logicians

Holistic and Holy Approach
Manager  Status: Persevering
Joined: 15 May 2013
Posts: 141
Location: India
GMAT Date: 08-02-2013
GPA: 3.7
WE: Consulting (Consulting)
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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.
SVP  Joined: 06 Sep 2013
Posts: 1506
Concentration: Finance
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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)
SVP  Joined: 06 Sep 2013
Posts: 1506
Concentration: Finance
Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

### Show Tags

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 495-40=455

Is this method valid?

Cheers!
J  Re: There are 12 cans in the refrigerator. 7 of them are red and   [#permalink] 24 Apr 2014, 14:39

Go to page    1   2    Next  [ 28 posts ]

Display posts from previous: Sort by

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