Lesson: The Fundamental Counting Principle

Comment on The Fundamental Counting Principle

Question at http://gmatclub.com/forum/a-palindromic-number-reads-the-same-forward-and-backward-214196.html is not clear to me. Could you please help me on this?

How many 6-digits number are Palindromic numbers? A Palindromic number reads the same forward and backward, example 12121.
A) 100
B) 610
C) 729
D) 900
E) 1000
gmat-admin's picture

I'm not a huge fan of that question because it first starts by talking about 6-digit numbers and then gives an example of a 5-digit number that works. However, my solution appears on that same page.
The basic idea is that, once we have selected the first 3 digits of the number, the last 3 digits automatically follow.
For example, if the first 3 digits are 356---, then (to be a palindrome), the last 3 digits must be ---652, so we get the number 356652.
Likewise, if the first 3 digits are 197---, then (to be a palindrome), the last 3 digits must be ---791, so we get the number 197791.

Brent, hi.
Why dont we take 0 in the account for the first digit?
gmat-admin's picture

Good question!

If we make 0 the first digit, then we no longer have a 6-digit number.
For example, 012210 appears to be a palindrome, but since 012210 = 12210, we we can't really say that 012210 is at 6-digit number.

Hey Brent! I love your videos so much. They are of great help to me!
My question is, how commonly is this topic tested on the GMAT? Is this a frequently tested topic? Is it advisable to skip this one topic out of all others or will it hurt my score significantly?

Thanks!
gmat-admin's picture

Glad you like the videos!

Counting questions aren't that common on the GMAT.

The number of questions YOU see on test day will depend on how well you're doing on the quantitative section. If you're doing really well, you MIGHT see 2 or 3 counting questions. If you're not doing well, you'll see FEWER counting questions.

If your prep time is limited, I suggest that you master the The Fundamental Counting Principle strategy, since it can be used for the majority of counting questions on the GMAT.

In this question below:
Team A and Team B are competing against each other in a game of tug-of-war. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups?
-Should I assume that 'decides to line up..." means that it is mandatory that the lineup be as such?
-I interpreted decided lineup in a way that the answer is 6! but it looks like there is restrictions here (answer being 3*3*2*2*1*1) but the language doesn't seem that clear. Thoughts? Is this an official GMAT lingo?
gmat-admin's picture

6! represents the number of ways to arrange 6 people WITHOUT any restrictions (6 ways to fill the first position, 5 ways to fill the second position , etc).

However, the restriction says that the first position must be a male. So, there are only 3 ways to fill that first position. Etc.

The question could be worded MUCH better.

Dear Brent,

I was trying to solve questions similar to:
The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

I am simply not able to wrap my head around any of approaches being used to solve this question. Neither the stages method explained in the GMATPrep videos nor the Manhattan slot method.

When I tried solving the it using the stages method, i went by solving as described below:

Stage 1: Car1 - 4 colors x 2 models
Car 2 - 3 colors x 2 models
Car 3 - 2 Colors x 2 models.

I am having a hard time understanding the repetition part of these kind of questions and always get stuck.
gmat-admin's picture

Check out my solution below, and please let me know if it makes sense: https://gmatclub.com/forum/the-carson-family-will-purchase-three-used-ca...

for the above problem -
cars and models ad colors.
The leaf diagram gives me 24.
3 cars * 2 models * 4 colors = 24.
What am I missing ?
gmat-admin's picture

First off, your solution does not take into account that the 3 cars must be 3 different colors.

To help better understand why your approach, please break your solution into steps/stages and tell me what you are doing at at point in the solution.

ASIDE: After answering a few questions involving the Fundamental Counting Principle, some students fall into the habit of simply finding the product of all of the numbers that appear in the question. This is not a good strategy.

Brent,

I did not understand your answer, could you help me with this below question?
http://www.beatthegmat.com/wonderful-p-c-ques-t271001-15.html

Thanks and regards
Pedro
gmat-admin's picture

Hi Brent,
for this question: How many possible ways can 3 girls (Rebecca, Kate, Ashley) go on a date with 3 boys (Peter, Kyle, Sam)?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 8

Assuming a date requires 1 boy + 1 girl
Stage 1 : selecting a boy. Can be done in 3 ways
Stage 2 : selecting a girl. Can be done in 3 ways

My answer is 9. However thats not one of the options. What gives ?
gmat-admin's picture

Question link: https://gmatclub.com/forum/how-many-possible-ways-can-3-girls-rebecca-ka...

The problem with your approach is that you are answering a different question.

You are answering the question "In how many ways can ONE girl (from Rebecca, Kate, Ashley) go on a date with ONE boy (from Peter, Kyle, Sam)?"

If that were the question, then the correct answer would, indeed, be 9.

However, in the original question, all 6 people are going on the date, and each person from one group gets paired with someone from the other group.

Does that help?

Cheers,
Brent

yes thanks a ton !

Hello for this question I completed this way and still got the same answer as you can you please confirm that my solution is right

A B C D E Number choices 1,2,3,4,5,6
(c is the middle digit )
A = 6 options
B = 5 options
C = 1 option ( 3 odd numbers in total if A was odd then B would have 2 options leaving C with 1 option)
D = 4
E = 3

6*5*1*4*3 = 360

Is my solution correct also?

gmat-admin's picture

I can help as soon as you tell me which question you're referring to :-)

Cheers,
Brent

Sorry,

thought I could paste the question link here. The question is:

How many different five-digit codes can be picked from the digits 1 through 6 if the middle digit must be odd and no two digits might be the same?

A) 420
B) 360
C) 180
D) 120
E) 60
gmat-admin's picture

Link: https://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-di...

When you use the Fundamental Counting Principle, you must be sure to deal with the most restrictive rule first. In this question, the most restrictive rule says that the middle digit must be odd.
If you don't deal with that first, then you end up with scenarios like you have where you say "C = 1 option ( 3 odd numbers in total if A was odd then B would have 2 options leaving C with 1 option)"

While you did, indeed, reach the same answer, I'm not entirely sure how you dealt with this mathematically.
I would avoid that approach in the future. Instead, be sure to deal with the most restrictive rule first.

Cheers,
Brent

Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and Y coordinates of P,Q and R are to be integers that satisfy the inequalitites -4≤ X≤ 5 and 6≤ y≤ 16. How many different triangles with these properties could be constructed?
A. 110
B. 1100
C. 9900
D. 10000
E. 12100
Hi Brent,
Could you please help me understanding the above question?

Thanks,
Isha Gaba
gmat-admin's picture

Hi Isha,

Here's my full solution: https://gmatclub.com/forum/right-triangle-pqr-is-to-be-constructed-in-th...

Cheers,
Brent

Dear Brent,

In this question:
Team A and Team B are competing against each other in a game of tug-of-war. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups?

the Total number of possible combinations (assuming different males and females) is 720.
What about calculating all the different combinations assuming that males and females as the same "object", should I use the Mississippi rule? (20?)

Thanks a lot.
gmat-admin's picture

Question link: https://gmatclub.com/forum/team-a-and-team-b-are-competing-against-each-...

Hi Dvd,
What steps did you take to get 720?

I'm assuming that you calculated 6! to get 720.
This means you're saying that there are 6 ways to choose someone to be 1st in line, and there are 5 ways to choose someone to be 2nd in line, etc...

HOWEVER, the 1st person in line must be a male. Since there are 3 males, there are 3 (not 6) ways to choose someone to be 1st in line.

When using the Fundamental Counting Principle, we must be sure to remind ourselves what we are trying to accomplish during EACH STAGE of our solution. I'm not saying that you need to write down each step (as I do in my solution here https://gmatclub.com/forum/team-a-and-team-b-are-competing-against-each-...), but you should definitely make sure you define (to yourself) what you're doing during each stage.

Does that help?

Cheers,
Brent

Hi Brent, I was thinking in general terms.
I want to know all the different combinations of these 3 men and 3 girls.
(MMMFFF, FFFMMM, MFMFMF, MMFMFF etc.)
is 720 correct?
gmat-admin's picture

If we're saying that the M's are identical, and the F's are identical, then we're just arranging the letters F, F, F, M, M, M
In this case, we can use the MISSISSIPPI rule (https://www.gmatprepnow.com/module/gmat-counting/video/785)
This will yield 20 arrangements

If we're saying that the M's are NOT identical, and the F's are NOT identical, but the males and females can be placed in ANY position (rather than alternating male/female), then the number of arrangements = 6!

Does that answer your question/

Cheers,
Brent

Yes, exactly!.

Thanks a lot!

Hi Brent, in the below question while I have understood the solution given but I am unclear of one particular option, which is 8! -1. Could you explain why this option is not the answer. I am failing to understand what 8! connotes in the given context.

https://gmatclub.com/forum/a-man-8-friends-whom-he-wants-to-invite-for-dinner-the-number-of-ways-239148.html
gmat-admin's picture

Question link: https://gmatclub.com/forum/a-man-8-friends-whom-he-wants-to-invite-for-d...

When answering questions using the Fundamental Counting Principle, we must be able to clearly define what we're doing at each stage (step) in the process.
So, before I can tell you what's wrong with the solution 8! - 1, I should ask you how you defined each stage of your solution.

That said, here's the logic (faulty logic) that some people might apply to the question to get that answer.

STAGE 1: Invite one of his friends
He has 8 friends, so this stage can be accomplished in 8 ways.

STAGE 2: Invite another of his friends
He has 7 friends remaining, so this stage can be accomplished in 7 ways.

STAGE 3: Invite another of his friends
He now has 6 friends remaining, so this stage can be accomplished in 7 ways.

I'll stop here.

At this point, we've already invited 3 friends, BUT the question says to invite AT LEAST 1 friend.
This means the man can invite only 1 person if he chooses. Or he can invite 2 people.

So, by the time we complete STAGE 3, we've already invited 3 people.
If we keep going to STAGE 8, we will have invited all 8 friends (which doesn't follow the question)

There's another issue too.

We're treating identical outcomes as different.

Let's say in stages 1 to 8, we invite the 8 friends in this order: A, C, G, D, F, E, H, B

Alternatively, we can also invite the 8 friends in this order: B, H, A, C, G, D, F, E

Under the above solution, these two outcomes are considered different, when they are the same.

Does that help?

Cheers,
Brent

Thanks Brent this helps! If I am understanding this correctly 8!-1 denotes 8! ways to invite 8 friends minus the possibility of not inviting 1 friend. Thus, inviting 7 out of 8 friends. Am i correct?
gmat-admin's picture

That is not correct.

8! represents the number of ways to arrange all 8 friends in a row. That's all.
Subtracting 1 is just subtracting 1.

Does that help?

Cheers,
Brent

Hi Brent,

Why can't we use combinations to solve this problem? Since Gas, Black, Standard is the same as Standard, Black, Gas the order is not important. Isn't this a fit case for Combinations - 8C3?
gmat-admin's picture

Be careful.

The 8 in 8C3 suggests that there are 8 things to choose from (gas, electric, hybrid, red, black, blue, automatic and standard).

However, choosing ANY 3 of these 8 options will have unintended results.

For example, what happens if the 3 items we choose are: red, black and blue?

This doesn't make any sense.

Since we must choose 1 item from EACH of the 3 categories, we can't just combine all 8 options and choose 3 of them.

Does that help?

Cheers,
Brent

"Does that help?"

Yeah, quite straight forward. Got to re wire my brain. Thanks very much Brent.

Hi Brent,

I was referring to your solution : https://gmatclub.com/forum/departments-a-b-and-c-have-10-employees-each-and-department-d-has-222084.html#p1710548

My thought process was that there were 2 steps to accomplish the task. First stage was to select 1 from A,B,C and 2nd was to select 2 from D. Hence I concluded that the solution should be 30C1 * 20C2

Could you please help me understand what is the flaw in my reasoning?

Thanks & Regards,
Abhirup
gmat-admin's picture

Solution link: https://gmatclub.com/forum/departments-a-b-and-c-have-10-employees-each-...

I believe you missed seeing a crucial word in the given information, which says "A task force is to be formed by selecting 1 employee from EACH of departments A, B, and C and 2 employees . . ."

Without the word EACH, the question suggests that we'll choose one employee from the TOTAL of 30 employees in departments A, B and C.
However, the question is really asking us to select 1 employee from department A, 1 employee from department B and 1 employee from department C (i.e., select 1 employee from EACH of departments A, B, and C)

Cheers,
Brent

Thanks Brent for pointing out. I reviewed the Q multiple times but this was one of those cases when one doesn't see whats obvious. Silly of me to make such a mistake at this stage of my prep.
gmat-admin's picture

I think Counting questions are the hardest to completely master.

https://gmatclub.com/forum/the-subsets-of-the-set-s-t-u-consisting-of-the-three-elements-s-t-143145.html
sir can we solve this by counting manually since max value is 16?
gmat-admin's picture

Questions link: https://gmatclub.com/forum/the-subsets-of-the-set-s-t-u-consisting-of-th...

Great idea, Manjot!
I went ahead with your idea and answered the question: https://gmatclub.com/forum/the-subsets-of-the-set-s-t-u-consisting-of-th...

Cheers,
Brent

https://gmatclub.com/forum/the-subsets-of-the-set-s-t-u-consisting-of-the-three-elements-s-t-143145.html

So what happens with the empty subset? Are we not taking that into account as that does not contain a t?

The way I approached the problem was as follows using FCP:

Number of s we can choose from = 4
Number of u we can choose from = 4
Since we don't want t in the subset = 1
Both w and x don't even make an appearance in the choices.

So using FCP = 4 x 4 = 16

Have I used this method right?
gmat-admin's picture

Question link: https://gmatclub.com/forum/the-subsets-of-the-set-s-t-u-consisting-of-th...

I think it may be a coincidence that your solution yielded the correct answer.
Can you tell me what you mean in your first stage when you write "Number of s we can choose from = 4"?
How are there 4 ways to chose an s?

Cheers,
Brent

https://gmatclub.com/forum/five-friends-ross-phoebe-chandler-joey-and-monica-decide-to-ha-236999.html

I don't understand why, in your approach, when you consider stage 2 for selecting a pizza your reasoning is that there are only 4 pizzas remaining for Ross. The question doesn't give us the restriction that everybody must have a different type of pizza.

Counting is a big weakness of mine and many people tend to do these problems via a slot method which has not really got me far. Your approach seems easy to follow so appreciate it if you can clarify the reasoning in this problem.
gmat-admin's picture

Question link: https://gmatclub.com/forum/five-friends-ross-phoebe-chandler-joey-and-mo...

Be careful. The question tells us that "no two friends will eat the same type of pizza"

You're not alone in your difficulties with Counting questions. Keep in mind that, on test day, you will likely see 0, 1 or maybe 2 counting questions. So, if you don't need a super huge Quant score, you can ignore Counting (and Probability) so that you can focus on topics that are tested more frequently.

Also keep in mind that we can often use the Listing and Counting strategy (https://www.gmatprepnow.com/module/gmat-counting/video/773) to solve the question OR to gain insights into the correct answer.

I hope that helps.

Cheers,
Brent

how are you suppose to draw out all the paths if there are many ( over 100?)
gmat-admin's picture

This video is meant to show how and why the Fundamental Counting Principle (FCP) works.
When answering counting questions, you need not create any such tree diagrams.

In the next video lesson (https://www.gmatprepnow.com/module/gmat-counting/video/776) I show how we can apply FCP to answer questions (without drawing any tree diagrams)

Try answering some of the questions in the Reinforcement Activities box, and you'll see what I mean.

Cheers,
Brent

https://gmatclub.com/forum/runners-v-w-x-y-and-z-are-competing-in-the-bayville-local-triathlo-255802.html

Hey!
Can you simplify this solution?
gmat-admin's picture

hello sir!
greetings of the day!

can you provide a better solution for this question-

https://gmatclub.com/forum/a-palindrome-is-a-number-that-reads-the-same-forward-and-129898.html

thanks,
tanya

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