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# If 3 different integers are randomly selected from the integers from 1

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If 3 different integers are randomly selected from the integers from 1 [#permalink]
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idkhur wrote:
This is what I did. If you do 12C3, you'll get 12!/9!3! and you'll get 220. Now you know that 220 will be in the denominator when calculating the probability. Looking at the answer choices, 44 is the only number that is a factor of 220, so that is the only possible answer.
C. 19/44

Perfect!!!!

I specifically created this question to illustrate the importance of calculating the denominator first (when using counting techniques to solve a probability question)
There are two reasons why you should calculate the denominator first:

1) The denominator is almost always easier to calculate than the numerator, and while calculating the denominator, you may gain some insight into how to calculate the numerator.
2) Once you know the denominator, you can use this to eliminate answer choices. So, even if you don't know the correct answer, you can still eliminate some answers and increase your chances of guessing correctly.

If you're lucky, you can eliminate 4 of the 5 answer choices, as you have done above!!

Cheers,
Brent
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If 3 different integers are randomly selected from the integers from 1 [#permalink]
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GMATPrepNow wrote:
If 3 different integers are randomly selected from the integers from 1 to 12 inclusive, what is the probability that a triangle can be constructed so that its 3 sides are the lengths of the 3 selected numbers?

A) 3/8
B) 7/18
C) 19/44
D) 39/88
E) 11/24

For this question, P(a triangle can be constructed with the 3 selected lengths) = (number of triangles with 3 lengths from 1 to 12 inclusive)/(TOTAL number of ways to select 3 numbers)

Let's calculate the denominator.

TOTAL number of ways to select 3 numbers
Since the order in which we select the 3 numbers doesn't matter, we can use COMBINATIONS.
We can select 3 numbers from 12 numbers in 12C3 ways.
12C3 = (12)(11)(10)/(3)(2)(1) = 220 ways

So, the correct answer will either be in the form k/220, OR some equivalent fraction in which the denominator is a FACTOR of 220.
For example, IF we calculate the numerator and get 110, then the answer = 110/220 = 1/2 (notice that 2 is a FACTOR of 220)
IF we calculate the numerator and get 15, then the answer = 15/220 = 3/44 (notice that 44 is a FACTOR of 220) And so on.

When we check the answer choices, we see that only one answer choice (C) has a denominator that's a FACTOR of 220.
So, C must be the correct answer.

We can answer the question without having to calculate the numerator (which is a time-consuming task)

On test day, it's unlikely that this technique will allow you to eliminate 4 answer choices. HOWEVER, if you're pressed for time, or you can't calculate the numerator, this technique may allow you to eliminate some of the answer choices and increase your likelihood of a correct guess.

Cheers,
Brent
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Re: If 3 different integers are randomly selected from the integers from 1 [#permalink]
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GMATPrepNow:
I have a doubt in the below Concept.
KEY CONCEPT: The longest side must be greater than the sum of the other two sides.

let's take an example from above.

The sample, 12,11,10. how Longest Side 12 is greater than sum of the other two sides.

In this case: 12 < 11+10 ? Could you please explain

Thanks
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Re: If 3 different integers are randomly selected from the integers from 1 [#permalink]
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ragav182 wrote:
GMATPrepNow:
I have a doubt in the below Concept.
KEY CONCEPT: The longest side must be greater than the sum of the other two sides.

let's take an example from above.

The sample, 12,11,10. how Longest Side 12 is greater than sum of the other two sides.

In this case: 12 < 11+10 ? Could you please explain

Thanks

Ooooops!!
That SHOULD have read "KEY CONCEPT: The longest side must be less than the sum of the other two sides."
I've edited my response accordingly.
Kudos for you!!

Cheers and thanks,
Brent
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Re: If 3 different integers are randomly selected from the integers from 1 [#permalink]
idkhur wrote:
This is what I did. If you do 12C3, you'll get 12!/9!3! and you'll get 220. Now you know that 220 will be in the denominator when calculating the probability. Looking at the answer choices, 44 is the only number that is a factor of 220, so that is the only possible answer.
C. 19/44

Used the same process. Wasn't able to get to the Pattern quick enough in which the 3 Integers chosen would make the successful side lengths.
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Re: If 3 different integers are randomly selected from the integers from 1 [#permalink]
Great question.

Used the same method: 12C3 = 95/220

Which led me to eliminate 4 answer choices and arrive at C.
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Re: If 3 different integers are randomly selected from the integers from 1 [#permalink]
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Re: If 3 different integers are randomly selected from the integers from 1 [#permalink]
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