# Question: Sum of 11

## Comment on Sum of 11

### Hi Brent! I would like to

Hi Brent! I would like to know if there is a different approach to know the number of pairs that sum 11. If the sets contained many numbers, how could we get to the numerator?

### 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

Hi Brent - Question about the denominator. Would it not be less than 40 since we have some duplicate pairs? ex. 2/2, 4/4, and 6/6?

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,

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

Hello! I have used combination to get the denominator 9C2, I don't understand why can't we do that.Thanks in advance :)

### 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

why can't we just multiple the individual probabilities together to get the cumulative probability?

(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

I realized what i did wrong it should be ( 4/5 * 1/8) if you pick a number from the first 5 numbers there is an 80% that it would be able to possibly sum to 11 given the numbers in set 2. The 1/8 probability stems from there only being on corresponding match to get a sum of 11.

### 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