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abhijit_sen
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We have 1 pair for each number between 1-6 inclusive. What 05 Jul 2008, 22:24
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We have 1 pair for each number between 1-6 inclusive. What is the probability that if 4 numbers are selected out of 12 then none of them belong to same pair?

Although all suggestions are welcome, I prefer If someone can solve it using combinatorics.

OA to follow later.



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We have 1 pair for each number between 1-6 inclusive. What Updated on: 07 Jul 2008, 22:51
Originally posted by code19 on 05 Jul 2008, 22:37.
Last edited by code19 on 07 Jul 2008, 22:51, edited 1 time in total.
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1st number can be any : 12/12
2nd number : 10/11 (11 left after the first one and 10 of them are eligible)
3rd number : 8/10 (10 left after the first and the second numbers, 8 of them are eligible)
4th number : 6/9 (9 left after 3 selections, 6 of them are eligible)


(12/12) (10/11) (8/10) (6/9) = 16/33
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We have 1 pair for each number between 1-6 inclusive. What 06 Jul 2008, 17:55
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Any more takers for it.
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mbawaters
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We have 1 pair for each number between 1-6 inclusive. What 15 Jul 2008, 18:10
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code19
1st number can be any : 12/12
2nd number : 10/11 (11 left after the first one and 10 of them are eligible)
3rd number : 8/10 (10 left after the first and the second numbers, 8 of them are eligible)
4th number : 6/9 (9 left after 3 selections, 6 of them are eligible)


(12/12) (10/11) (8/10) (6/9) = 16/33
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deto code19!

Whats the OA?
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alpha_plus_gamma
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We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 18:15
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abhijit_sen
We have 1 pair for each number between 1-6 inclusive. What is the probability that if 4 numbers are selected out of 12 then none of them belong to same pair?

Although all suggestions are welcome, I prefer If someone can solve it using combinatorics.

OA to follow later.
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I don't think my answer is right but I tried to solve it by combinatorics.

heres my approach

4 numbers out of 12 = 12C4 = 495
we have six pairs of numbers,

possible of getting 4 numbers such that they belong to pair
= selecting 2 pairs of numbers out of 6 pairs
= 6C1 * 5C1
= 30

probability of getting 4 numbers that belong to pair = 30/495

so,
probabilty of selecting 4 numbers such that none of them belong to same pair
= 1 - probability of getting 4 numbers that belong to pair
= 1 - 30/495
= 465/495
= 31/33

I don't think this is right answer. I have messed up somewhere in between.
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We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 18:50
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total number C(12,4)

number of no pairs: first choose 4 out of 6 pairs C(6,4) then for each pair there are two options 2^4

probability = C(6,4)*2^4/C(12,4)=16*(6!/(4!*2!))/(12!/(4!*8!)=8*6!/(12!/8!)=8*6!/(12*11*10*9)=8*6*5*4*3*2/(12*11*10*9)=8*3*2/(11*9)=16/33
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We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 18:52
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maratikus
total number C(12,4)

number of no pairs: first choose 4 out of 6 pairs C(6,4) then for each pair there are two options 2^4

probability = C(6,4)*2^4/C(12,4)=16*(6!/(4!*2!))/(12!/(4!*8!)=8*6!/(12!/8!)=8*6!/(12*11*10*9)=8*6*5*4*3*2/(12*11*10*9)=8*3*2/(11*9)=16/33
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can you please explain why you choose 4 out of 6 pairs? we have to choose 4 numbers right? so 2 pairs should be selected.
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We have 1 pair for each number between 1-6 inclusive. What 16 Jul 2008, 20:05
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maratikus
total number C(12,4)

number of no pairs: first choose 4 out of 6 pairs C(6,4) then for each pair there are two options 2^4

probability = C(6,4)*2^4/C(12,4)=16*(6!/(4!*2!))/(12!/(4!*8!)=8*6!/(12!/8!)=8*6!/(12*11*10*9)=8*6*5*4*3*2/(12*11*10*9)=8*3*2/(11*9)=16/33
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Right on point. I was waiting for something like this. You hit the bulls eye.



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