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

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?

'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

\(12C4 - 7C4 - 5C4 = 495 - 35 -5 = 455\)

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.

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.

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.
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its 7C6*5C2+7C5*5C3+7C4*5C4=70+210+175=455

My take is D

Any combination of 4 cans – combination without red cans – combination without blue cans = 12C4 – 7C4 – 5C4 = 455

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

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

\(C^n_k=\frac{n!}{k!(n-k)!}\)

Check here: math-combinatorics-87345.html
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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 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.
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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.
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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
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Yep, did it the other way around

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

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
Answer is thus D

Is this method valid?

Cheers!
J