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

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

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New post 30 May 2014, 04:23
I am not sure why in all above solutions cans are considered distinct. Question doesn't say explicitly that cans are distinct or identical.

If cans are identical then:
6 red, 2 blue (1 red, 3 blue are left) --> RRRRRR BB can be taken out from fridge in ways = \(\frac{4!}{6!*2!}\)
5 red, 3 blue (2 red, 3 blue are left) --> RRRRR BBB can be taken out from fridge in ways = \(\frac{4!}{5!*3!}\)
4 red, 4 blue (3 red, 1 blue are left) --> RRRR BBBB can be taken out from fridge in ways = \(\frac{4!}{4!*4!}\)

If all cans are distinct then :
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) --> \(C^4_5*C^4_7=175\);

I solved this problem second time and solved it same way I solved last time.
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There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

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New post 26 Jan 2016, 13:56
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.
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Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

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New post 27 Jan 2016, 05:30
The total number of ways in which 8 cans can be chosen from 12 cans is 12C8.

In order to find out the ways in which at least 1 can from Blue and at least one can from Red is left, would require us to count all possible scenarios in which 2 Red Ball and 1 Blue Ball remain, then, 2 Red Ball and 2 Blue Ball remain. . . . .

In order to shorten our no. of steps,

we can simply find out the no. of ways we can subtract the number of times there are no red ball and the number of times there are no blue balls.

Total no. of ways to choose 8 balls from 12 is given as 12C8.

Total no. of ways when no red balls remain is 7C7 + 5C1
Total no. of ways when no blue balls remain is 5C5 + 7C3.

Combine these total unacceptable ways = 1*5 + 1* 7*5
= 40
Total number of ways = 12C8 = 495

Total number of acceptable ways = 495-40 = 455
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Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

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New post 01 Feb 2016, 00:52
2
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!.
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Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

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New post 10 Feb 2016, 04:05
VeritasPrepKarishma wrote:
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!.


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

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New post 17 Apr 2017, 20:46
Option D

Total cans = 12, Red = 7 & Blue = 5.

Total no. of ways to pick 8 cans = 12C8 = 495

No. of ways to pick all Red cans or all Blue cans = 7C7.5C1 + 7C3.5C5 = 40

Answer = 495 - 40 = 455
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Re: There are 12 cans in the refrigerator. 7 of them are red and  [#permalink]

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New post 17 Apr 2017, 20:59
My take is D

4 cans will remain in the fridge. Combinations-(3,1),(1,3),(2,2)
7C1*5C3 + 7C3*5C1 + 7C2*5C2


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

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Re: There are 12 cans in the refrigerator. 7 of them are red and   [#permalink] 10 Apr 2019, 12:44

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