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Two integers will be randomly selected from sets A and B,

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by AAPL » Tue May 08, 2018 4:39 am

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Answers

A

B

C

D

E

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Difficulty—

A = {2, 3, 5, 7, 11}
B = {2, 4, 6, 13}

Two integers will be randomly selected from sets A and B, one integer from set A and one from set B, and then multiplied together. How many different products can be obtained?

A. 15
B. 16
C. 19
D. 20
E. 36

The OA is C.

It can be easily done by making pairs, but can somebody tell me how to do it using combinations or some other short-cut? Thanks!
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Two integers will be randomly selected from sets A and B,

by Brent@GMATPrepNow » Tue May 08, 2018 5:35 am
AAPL wrote:A = {2, 3, 5, 7, 11}
B = {2, 4, 6, 13}

Two integers will be randomly selected from sets A and B, one integer from set A and one from set B, and then multiplied together. How many different products can be obtained?

A. 15
B. 16
C. 19
D. 20
E. 36
Take the task of creating products and break it into stages.

Stage 1: Select a number from set A
There are 5 numbers to choose from, so we can complete stage 1 in 5 ways

Stage 2: Select a number from set B
There are 4 numbers to choose from, so we can complete stage 2 in 4 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create our products) in (5)(4) ways (= 20 ways)

HOWEVER, we need to recognize that there may be some DUPLICATE outcomes among the 20 outcomes.

For example, among the 20 outcomes, we get a product of 12 by selecting the 2 from set A and the 6 from set B. However, we can also get a product of 12 by selecting the 3 from set A and the 4 from set B.
So, we have selected the same product (of 12) TWICE. So, we must subtract one of those outcomes from our total of 20 outcomes to get 19 DIFFERENT products.

Are there any other products that we have counted twice?
No there aren't. So, we're done.

Answer: C
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
- https://www.beatthegmat.com/counting-pr ... 44302.html
- https://www.beatthegmat.com/picking-a-5 ... 73110.html
- https://www.beatthegmat.com/permutation ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatori ... 73194.html
- https://www.beatthegmat.com/arabian-hor ... 50703.html
- https://www.beatthegmat.com/sub-sets-pr ... 73337.html
- https://www.beatthegmat.com/combinatori ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-se ... 71047.html
- https://www.beatthegmat.com/combinatori ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/permutation ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Wed May 09, 2018 3:53 pm
AAPL wrote:A = {2, 3, 5, 7, 11}
B = {2, 4, 6, 13}

Two integers will be randomly selected from sets A and B, one integer from set A and one from set B, and then multiplied together. How many different products can be obtained?

A. 15
B. 16
C. 19
D. 20
E. 36
Since there are 5 numbers in set A and 4 numbers in set B, the maximum number of distinct products we can have is 5 x 4 = 20 if there are no duplicates. Let's now determine the number of duplicates.

While determining the duplicates, we should note that the numbers 5, 7 and 11 in A are prime numbers and none of these numbers are also in B; therefore no product containing 5, 7 or 11 from set A can be a duplicate. Similarly, the number 13 in B is a prime number that is not contained in A; therefore, any product containing 13 will not be a duplicate, either. To make our job easier, we should look for duplicates in the products formed by multiplying 2 or 3 from set A by 2, 4 or 6 from set B.

2 x 6 = 3 x 4 = 12

There is only 1 duplicate. Therefore, the number of distinct products is 20 - 1 = 19.

Answer: C

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