Lesson: Fundamental Counting Principle Example

Comment on Fundamental Counting Principle Example

If we have 2 ways of sitting the girls in chair no. 1, don't we have 2 ways of sitting the girls in chair no. 5 too. Likewise, 3 ways for 2, 3, 4?. Just trying to look at the flip side too. Thanks.
greenlight-admin's picture

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/qotd-19-the-5-letters-in-the-list-g-h-i-j-k-are-to-be-2652.html
.........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?
greenlight-admin's picture

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 inconvenience. I had for some weird
reason thought that your answer to the ques is 6(maybe I was
hallucinating). Thank you so much for the quick reply.
greenlight-admin's picture

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

Cheers,
Brent

What do you meab by restrictive stage
greenlight-admin's picture

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 wondering why my I've not been getting reponses to my question on your module and also the daily prep questions. I thought someting was wrong with my account till i checked spam and saw your responses there. Is there something you can do about it? And my daily prep questions aren't there also
greenlight-admin's picture

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 guides for GMAT that we can use the factorial for a problem like this. by treating the entire total items ( Girls plus Boys) as 5! and then take 5! - (4!)*2? Please let me know if this would be an appropriate strategy and where I am going wrong? Thanks.
greenlight-admin's picture

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.

Add a comment

Have a question about this video?

Post your question in the Comment section below, and I’ll answer it as fast as humanly possible.

Change Playback Speed

You have the option of watching the videos at various speeds (25% faster, 50% faster, etc). To change the playback speed, click the settings icon on the right side of the video status bar.

Let me Know

Have a suggestion to make the course even better? Email me today!

Free “Question of the Day” emails!