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## Comment on

Sum of 11## Hi Brent! I would like to

Thanks in advance!

## Hi Pau,

Hi Pau,

This question requires some brute force to count the possible pairs.

The GMAT test makers are very reasonable. So they wouldn't require you to perform the same tedious steps with tons of numbers. If they did provide large sets of numbers, then there would most likely be a pattern that you could recognize to save yourself time.

Cheers,

Brent

## Many thanks :)

## Hi Brent - Question about the

Thanks,

Jon

## Great question!

Great question!

It turns out that we haven't counted any possible outcomes more than once. Here's why:

When counting the total number of possible outcomes, stage 1 was selecting from the set {1,2,4,6,7}, and stage 2 was selecting from the set {2,3,4,5,6,7,8,9}

For example, this means there's only one way to get the outcome 2/2.

The first 2 is from stage 1, and the second 2 is from stage 2.

Cheers,

Brent

## Hi Brent,

I am a bit confused about calculating the total no: of outcomes in the denominator.

How is 13C2 incorrect? There are 13 distinct no:s and we are selecting 2 no:s? Do the two sets make them independent so that we have to calculate the selection separately?

Is it not the same as selecting 2 no:s that add to 11 from a set of (2,3,4,5,6,7,8,9,1,2,4,6,7}.

Am I missing an important concept here?

Thank you!

## If we set up our solution so

If we set up our solution so that the denominator equals 13C2, then we are counting some outcomes that shouldn't be counted.

For example, one of the 13C2 outcomes would be selecting a 3 and a 5. However, the question tells us to select ONE number from EACH set, and the 3 and the 5 only appear in the second set.

Cheers,

Brent

## Hello! I have used

## I'm assuming that you're

I'm assuming that you're choosing 2 values from the set {1,2,3,4,5,6,7,8,9}

This doesn't work, because it misses some possible outcomes.

Notice that your approach doesn't allow us to have 2 MATCHING numbers.

For example, we could choose a 6 from the first set, and also choose a 6 from the second set.

However, if you use combinations to choose 2 values from the set {1,2,3,4,5,6,7,8,9}, the approach will not count getting a 6 from each set.

Does that help?

Cheers,

Brent

## why can't we just multiple

(4/5 * 4/8)

## In your solution, what do 4/5

In your solution, what do 4/5 and 4/8 represent?

## I realized what i did wrong

## That's perfect.

That's perfect.

Aside: this equation appears very early in the Probability module, before we cover probability rules.

This is why I solved the question using only counting methods.

Cheers,

Bretn

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