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If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6

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Bunuel
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If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   06 Oct 2017, 01:03
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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 square of an integer?

A. 1/2
B. 1/3
C. 1/5
D. 1/9
E. 1/30
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niks18
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If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   06 Oct 2017, 07:04
Bunuel 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 square of an integer?

A. 1/2
B. 1/3
C. 1/5
D. 1/9
E. 1/30


Total no of ways of selecting two numbers out of 6 numbers = \(6_C_2\) \(= 15\)

Possible two number that satisfy the given condition are = {1,3}, {3,6} & {4,5} = 3

Hence Probability \(= \frac{3}{15}=\frac{1}{5}\)

Option C
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Re: If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   06 Oct 2017, 16:28
niks18 wrote:
Bunuel 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 square of an integer?

A. 1/2
B. 1/3
C. 1/5
D. 1/9
E. 1/30


Total no of ways of selecting two numbers out of 6 numbers = \(6_C_2\) \(= 15\)

Possible two number that satisfy the given condition are = {1,3}, {3,6} & {4,5} = 3

Hence Probability \(= \frac{3}{15}=\frac{1}{5}\)

Option C


Hello,

can you explain why you did this?

Number of ways,

then the possibilities fulfilling the condition. Any reosurce on counting methods would help
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Re: If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   06 Oct 2017, 19:17
JulesGaire wrote:
niks18 wrote:
Bunuel 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 square of an integer?

A. 1/2
B. 1/3
C. 1/5
D. 1/9
E. 1/30


Total no of ways of selecting two numbers out of 6 numbers = \(6_C_2\) \(= 15\)

Possible two number that satisfy the given condition are = {1,3}, {3,6} & {4,5} = 3

Hence Probability \(= \frac{3}{15}=\frac{1}{5}\)

Option C


Hello,

can you explain why you did this?

Number of ways,

then the possibilities fulfilling the condition. Any reosurce on counting methods would help


Hi JulesGaire

Sorry but I could not understand your query properly. It would be great if you could let me know what is the specific problem you are facing.

If your query is related to the basics of Permutation & Combination or Probability then below resources from GMAT club & Bunuel will help you a lot

1. GMAT Club Math PDF book - https://gmatclub.com/forum/gmat-math-bo ... 30609.html
2. Official Guide Forum - https://gmatclub.com/forum/gmac-officia ... 40610.html
3. Quant Megathread - https://gmatclub.com/forum/ultimate-gma ... 44512.html
4. Advanced Topic on Probability & P&C - https://gmatclub.com/forum/advanced-top ... 01361.html
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If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   07 Oct 2017, 09:24
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niks18 wrote:
Bunuel 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 square of an integer?

A. 1/2
B. 1/3
C. 1/5
D. 1/9
E. 1/30


Total no of ways of selecting two numbers out of 6 numbers = \(6_C_2\) \(= 15\)

Possible two number that satisfy the given condition are = {1,3}, {3,6} & {4,5} = 3

Hence Probability \(= \frac{3}{15}=\frac{1}{5}\)

Option C



You got the answer right but shouldn't it be like below?

Number of ways you can select 2 values from 6 where order of selected 2 matter --> 6!/(6-2)! = 30
The squares possible within the range given here is 4 and 9

So possibilities are: {1,3},{3,1},{3,6},{6,3},{4,5},{5,4} --> 6 options.

So probability = Number of required outcomes / Total number of possible outcomes = 6/30=1/5
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Re: If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   07 Oct 2017, 09:32
rajudantuluri wrote:
niks18 wrote:
Bunuel 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 square of an integer?

A. 1/2
B. 1/3
C. 1/5
D. 1/9
E. 1/30


Total no of ways of selecting two numbers out of 6 numbers = \(6_C_2\) \(= 15\)

Possible two number that satisfy the given condition are = {1,3}, {3,6} & {4,5} = 3

Hence Probability \(= \frac{3}{15}=\frac{1}{5}\)

Option C



You got the answer right but shouldn't it be like below?

Number of ways you can select 2 values from 6 where order of selected 2 matter --> 6!/(6-2)! = 30
The squares possible within the range given here is 4 and 9

So possibilities are: {1,3},{3,1},{3,6},{6,3},{4,5},{5,4} --> 6 options.

So probability = Number of required outcomes / Total number of possible outcomes = 6/30=1/5


Hi rajudantuluri

Can you clarify why order should matter here? the question does not mention that the selected numbers are arranged in any manner (row/circular formation etc.). Hence in my opinion Permutation is not required here.
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Re: If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   07 Oct 2017, 10:54
Because this asks for a probability when you pick two numbers.

Let’s says you have a dice you can roll twice and we’re asking what the probability is to get a 9 with 6&3.

Won’t you say there are two ways?

1st option:
First roll you get a 3 and second roll you get a 6.

2nd option:
First roll you get a 6 and second roll you get a 3.

In probability shouldn’t we consider both these options.

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Re: If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   07 Oct 2017, 11:12
rajudantuluri wrote:
Because this asks for a probability when you pick two numbers.

Let’s says you have a dice you can roll twice and we’re asking what the probability is to get a 9 with 6&3.

Won’t you say there are two ways?

1st option:
First roll you get a 3 and second roll you get a 6.

2nd option:
First roll you get a 6 and second roll you get a 3.

In probability shouldn’t we consider both these options.

Posted from my mobile device


Hi rajudantuluri

I agree with a dice the possibility that you mentioned can arise. But here we have a set.
Imagine a bucket with 6 cubes numbered 1 to 6. you need to pick two cubes. once a cube is picked it is not put back in the bucket. So once you pick out 1 & 3, then the bucket will not contain any other 1 & 3 to be picked up again. Hence repetition is not possible.

Hi Bunuel
Need your expertise to clarify whether repetition is allowed here or not i.e to say is {1,3} & {3,1} different possibilities or the same?
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Re: If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6 [#permalink] New post   25 Feb 2021, 12:11
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