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If two different numbers are to be selected from set {1, 2,

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If two different numbers are to be selected from set {1, 2, [#permalink]

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29 Jul 2008, 09:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

. If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6}, what is the probability that the sum of two numbers is a perfect square number?
A. 1/2
B. 1/3
C. 1/9
D. 2/9
E. 4/9

here what confuses me is , in the total poss cases, do we take ,for eg 1,3 and also along with that 3,1 or should they be considered only once?
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Re: probability [#permalink]

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29 Jul 2008, 09:51
Here perfect squares can be either 4 or 9.

4 = 3+1 or 1+3 (2 ways)
9 = 3+6 or 6 +3 (2 ways)
= total 4 ways

Total ways of selecting to numbers = 6 *6 = 36 ways

Required probabilty = 4/36 = 1/9
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Re: probability [#permalink]

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29 Jul 2008, 09:57
what abot 4,5 and 5,4?
also total ways of selecting two nos. from 6 is 6c2 rite?
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Re: probability [#permalink]

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29 Jul 2008, 10:02
We have 15 possible option and 5 good option. Probability=5/15=1/3
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Re: probability [#permalink]

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29 Jul 2008, 10:10
how 5 good option? it shud be 6 rite?
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Re: probability [#permalink]

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29 Jul 2008, 11:09
I think we have 5 good options: (1,3) (2,6) (3,5) (3,6) (4,5)

I hpe these are the perfect squere numbers: 4, 8, 9, ...

I am not sure, may be I am not correct.
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Re: probability [#permalink]

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29 Jul 2008, 16:39
arjtryarjtry wrote:
. If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6}, what is the probability that the sum of two numbers is a perfect square number?
A. 1/2
B. 1/3
C. 1/9
D. 2/9
E. 4/9

here what confuses me is , in the total poss cases, do we take ,for eg 1,3 and also along with that 3,1 or should they be considered only once?

2^2 = 4 and 3^2 = 9 are the only perfect squares we can get as a sum of two numbers from the set. Note that 8 is not a perfect square (8 is a cube, however).

Now, we can either assume order matters, or assume it doesn't- we just need to be consistent. If order matters, there are 6*5 = 30 different pairs of numbers we can choose. Again, if order matters, we can get squares in the following ways:
1,3
3,1
3,6
4,5
5,4
6,3

So the probability should be 6/30 = 1/5. That isn't among the answer choices- where is the question from?
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Re: probability [#permalink]

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29 Jul 2008, 23:14
number of ways that you can select 2 numbers out of 6 would be 6C2=15 ways
favourable outcomes are {1,3}, {4,5}, & {3,6} = 3 ways

hence the probability would be 3/15=1/5 (well it is not one of the options!!!)
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Re: probability [#permalink]

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29 Jul 2008, 23:18
hi0parag,

Thank you for correcting my mistake.
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Re: probability [#permalink]

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30 Jul 2008, 02:26
Wrong Question.

The answer is 1/5
Re: probability   [#permalink] 30 Jul 2008, 02:26
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