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

Question

There are 12 different 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

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

Bunuel · Accepted Answer

Obviously not.

There are 12 different 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.

swatirpr · Answer

'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

BarneyStinson · Answer

Logic wise does this logic work?

Total number of ways to selecting 8 cans from 12 cans MINUS

12C8 -   = 495 - 40 = 455.

D is the answer, what is the OA?

Bunuel · Answer

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.

PiyushK · Answer

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 ?

StormedBrain · Answer

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".

Bunuel · Answer

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.

StormedBrain · Answer

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

Bunuel · Answer

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.

SVaidyaraman · Answer

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

ramannanda9 · Answer

Yep, did it the other way around

7c6*5c2+7c5*5c3+7c4*5c4=455  Hence,D

leve · Answer

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.

KarishmaB · Answer

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!.

leve · Answer

Thank you Karishma. Yes you are right. The question is purely asking the number of ways of sellecting (removing) distinct cans. -Not arranging them after sellecting them. It is so straight forward now. Thanks again.

PrameshLuitel123 · Answer

I got the answer by doing it the other way around where I calculated the remaining 4 cans in the refrigirator, 5c1x7c3 + 5c3x7c1 + 5c2x7c2 and got to 455 is the logic right or I just got lucky with this question?

Bunuel · Answer

We are removing 8 cans, so the sum of red and blue removed must be 8. The direct way this would be:

7C6 * 5C2 + 7C5 * 5C3 + 7C4 * 5C4 = 455

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

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

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