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## Comment on

Fundamental Counting Principle Example## If we have 2 ways of sitting

## Once we COMPLETE stage 1

Once we COMPLETE stage 1 (placing a girl is chair 1), then there is only ONE girl remaining.

So, when we start working on stage 2 (placing a girl in chair 5), we have only ONE way to complete this stage.

The same applies to seating the boys.

Does that help?

## https://greprepclub.com/forum

.........I actually managed to derive 12 re-arrangements for this question through listing method which are....

1)H,I,G,J,K

2)H,J,G,I,K

3)H,K,G,I,J

4)H,I,G,K,J

5)H,J,G,K,I

6)H,K,G,J,I

7)J,K,G,I,H

8)J,I,G,K,H

9)I,J,G,K,H

10)I,K,G,J,H

11)K,I,G,J,H

12)K,J,G,I,H

Could you kindly help me with this?

## Question link: https:/

Question link: https://greprepclub.com/forum/qotd-19-the-5-letters-in-the-list-g-h-i-j-...

Your list of possible outcomes is perfect!

I'm not sure what else to add to your approach.

Here's my full solution: https://greprepclub.com/forum/qotd-19-the-5-letters-in-the-list-g-h-i-j-...

Cheers,

Brent

## Extremely sorry for the

reason thought that your answer to the ques is 6(maybe I was

hallucinating). Thank you so much for the quick reply.

## Not to worry; it's never an

Not to worry; it's never an inconvenience :-)

Cheers,

Brent

## What do you meab by

## The most restrictive stage is

The most restrictive stage is the stage that has the most rigid rules.

For example, let's say we must create a positive 3-digit integer in which all 3 digits must be different, AND the middle digit must be a 6 or a 7.

So, when we start listing different stages, we should begin with the one that has the most rigid rules.

In the case with the above question, we should start with the stage in which we select the middle digit (since it MUST be either 6 or 7).

Here's why:

Let's say that we didn't start with the most restrictive stage, and we set up our solution as follows:

STAGE 1: Select first digit

STAGE 2: Select middle digit

STAGE 3: Select last digit

STAGE 1 can be completed in 9 ways (choose 1,2,3,4,5,6,7,8 or 9)

STAGE 2 can be completed in 2 ways (choose 6 or 7)

STAGE 3 can be completed in 8 ways (choose any digit other than the two digits selected in STAGES 1 and 2)

This strategy is incorrect, because it allows us to select a 6 or 7 in STAGE 1.

So, if we did select a 6 in STAGE 1, then we can complete STAGE 2 in only ONE way (as opposed to 2 ways)

Conversely, if we did NOT select a 6 or a 7 in STAGE 1, then we can complete STAGE 2 in only TWO ways.

This creates a problem.

For this reason, we should start with the most restrictive stage.

Cheers,

Brent

## Hello Brent. I've been

## For some reason, your email

For some reason, your email software has categorized emails from @greenlighttestprep.com as spam.

You should be able to go into your email software and direct it to stop categorizing emails from that address as spam.

Cheers,

Brent

## I remember reading in the MP

## When it comes to using the

When it comes to using the Fundamental Counting Principle (FCP), it's important to understand what each value (stage) stands for. Otherwise, a formula like the one you're suggesting can be easily rendered ineffective if one small change is made to the question.

The formula that you're referring to (which yields the expression 5! - 2*4!) is the formula for determining the number of ways to seat x people, if two of the people cannot sit next to each other. In these instances, the correct answer will be x! - (2)(n-1)!

So, if the original question asked, "In how many ways can we seat all five children so that the two girls are NOT sitting next to each other?" then the correct answer would be 5! - (4!)*2

However, the question tells us that the the two girls must sit on the END CHAIRS. So, in this case, the formula does not apply.

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