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## Comment on

Distributing Coins## Can we solve this sum through

## As I mention in the video,

As I mention in the video, this question is out of scope for the GMAT. I wanted to use it to demonstrate the utility of listing and counting possible outcomes.

A "mathematical" approach would involve a technique known as "partitioning."

## I thought that we need to

## If you check the list, you'll

If you check the list, you'll see that {6,0,0}, {0,6,0} and {0,0,6} are already listed separately in the list of 28 outcomes.

## Hello, Didn't understand why

## To see what we can't use the

To see why we can't use the FCP, let's see what happens when we start.

STAGE 1: Give Alex some coins

We can give Alex 0, 1, 2, 3, 4, 5 or 6 of the coins.

So, we can complete stage 1 in 7 ways

STAGE 2: Give Bea some coins

In how many ways can we complete stage 2?

It depends on how many of the 6 coins we gave to Alex.

- If we gave 0 coins to Alex, then there are 6 coins remaining, which means we can give Bea 0, 1, 2, 3, 4, 5, or 6 coins.

- If we gave 1 coin to Alex, then there are 5 coins remaining, which means we can give Bea 0, 1, 2, 3, 4, or 5 coins.

- If we gave 2 coins to Alex, then there are 4 coins remaining, which means we can give Bea 0, 1, 2, 3 or 4 coins.

- If we gave 3 coins to Alex, then there are 3 coins remaining, which means we can give Bea 0, 1, 2 or 3 coins.

Etc...

So, there's no way to determine the number of ways to complete stage 2. The same goes for stage 3.

So, we need a different approach.

## Thanks, understood.

## Thank you! I had the same

## Hi Brent

Can we solve this question using fundamental counting principle?

Regards

Neha

## To my knowledge, there's no

To my knowledge, there's no nice way to solve this question using the Fundamental Counting Principle.

There is a technique, called the Separator Method, that we can use, but it's beyond the scope of the GMAT.

Cheers,

Brent

## in probability theory from my

## That's correct. However, the

That's correct. However, the GMAT does not require us to know this rule.

## Hi Brent, not sure why is not

## For this question, we are

For this question, we are simply listing and counting all of the possible outcomes.

So there's no need to introduce any other operations.

You are correct about the FCP. Since we are not using the FCP to answer the question, there's no need to multiply.

## Get it, thanks Brent

## Hi,

Since the question mentioned that there are "6 identical coins be distributed". Can I understand identical as that these coins look different, so the question will be even more complicated.

## I'm not sure what you're

I'm not sure what you're asking.

If all 6 coins are identical, then, for example, there's exactly 1 way Alex, Bea and Chad can receive 2 coins each.

If we say the 6 coins are all different, then there would be 90 ways Alex, Bea and Chad can receive 2 coins each.

So, yes, if the 6 coins are different, question becomes more complex, but that's not what the question is asking.