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## Comment on

The Fundamental Counting Principle## Greetings to you, Tanya!

Greetings to you, Tanya!

Here's my full solution: https://gmatclub.com/forum/a-palindrome-is-a-number-that-reads-the-same-...

Cheers, Brent

## https://gmatclub.com/forum/a

Hi Brent,

Could you help in this question please ?

Thanks,

Karaan

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/a-certain-company-assigns-employees-to-office...

## Hi Brent. Good Morning.

I couldn't understand your explanation at https://gmatclub.com/forum/the-subsets-of-the-set-s-t-u-consisting-of-the-three-elements-s-t-143145.html#p2185734,

Can you please simplify it?

## Solution link: https:/

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

The linked question above is very similar to the following video question in which we can customize a bike with 5 optional features: https://www.gmatprepnow.com/module/gmat-counting/video/794

For each of the 5 features in the bike question, we have the option of having the feature or not having the feature on the bike. In other words, for each feature, we have two options: Have on the bike or don't have on the bike.

The same approach applies to the original question where we must find subsets of the set {s, t, u, w, x}.

For 4 of the 5 elements in the set {s, u, w, x}, we have two options:

1) include the element in the subset

2) don't include the element in the subset

For the element t, we have only one option:

1) don't include the element in the subset

So, for EACH of the four elements, s, u, w, x, there are 2 options (have in the subset, or don't have in the subset)

For the element t, there's only 1 option (don't have in the subset)

So the total number of subsets = (2)(2)(2)(2)(1) = 16

Does that help?

## Hi Brent. Good Morning.

Can you please explain the solution of question - https://gmatclub.com/forum/for-dinner-at-a-restaurant-there-are-x-choices-of-appetizers-y-219917.html?

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/for-dinner-at-a-restaurant-there-are-x-choice...

## Hi Brent,

https://gmatclub.com/forum/a-palindrome-is-a-number-that-reads-the-same-forward-and-bac-161167.html#p1275137

For the question above, if we were asked to find the even number of palindrome, it would still be 50, right? If there wasn't any limitation on even or odd, then the total number of palindromes would be 100?

Thanks

## The link you provided doesn't

The link you provided doesn't match up with your question. Is the link correct?

## Apologies, it was a similar

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

## Question link: https:/

Question link: https://gmatclub.com/forum/a-palindrome-is-a-number-that-reads-the-same-...

If we were asked to find the number of 4-digit EVEN numbers that are palindromes, the correct answer would be 40.

The reason for this is that the units digit can't be 0, since the number would no longer be a 4-digit number.

Here's what I mean:

The number 4114 is an acceptable number, but the number 0110 is not an acceptable number, because 0110 is really just a weird way to write 110, which is a 3-digit number.

## Hi Brent,

In this question: https://gmatclub.com/forum/a-certain-company-assigns-employees-to-offices-in-such-a-way-88936.html

I was confused about what I need to choose as a stage - if I should assume the stages as no. of ways in which an employee will get assigned to the office, OR, if I should take Office 1 and Office 2 as the base to create the stages.

What should be the best approach to learn how we should go about creating the stages?

## Question link: https:/

Question link: https://gmatclub.com/forum/a-certain-company-assigns-employees-to-office...

As you've noted, there are two possible options with regard to the stages:

1) assign each person to an office

2) assign each office to a person

There's a serious flaw with the option #2, which I'll illustrate by attempting to solve the question.

Let A, B and C be the 3 employees

Let X and Y be the two offices

STAGE 1: Select the employees to work in office X

The number of ways to complete stage 1 will depend on how many employees will work in office X.

If 0 employees are to work in office X, then stage 1 can be completed in 1 way

If 1 employee is to work in office X, then stage 1 can be completed in 3C1 ways

If 2 employees are to work in office X, then stage 1 can be completed in 3C2 ways

If 3 employees are to work in office X, then stage 1 can be completed in 3C3 ways

As you can see, option #2 falls apart pretty quickly since each office can hold 0, 1, 2 or 3 employees.

Compare this strategy with option #1.

STAGE 1: Select an office for employee A to work in

There are two ways to accomplish Stage 1 (employee A can work in office X or office Y)

STAGE 2: Select an office for employee b to work in

There are two ways to accomplish Stage 2 (employee b can work in office X or office Y)

And so on...

Option #1 works because each employee must be assigned an office.

Option #2 doesn't work because each office does NOT need to be assigned an employee.

## Oh yes got it. So basically

## Exactly.

Exactly.

## Hi Brent,

Can you please explain this question?

There are 5 couples. If they sit on 10 chairs around a round table such that each couple sits side by side, how many possible cases are there?

A. 256

B. 512

C. 768

D. 1,024

E. 1,080

The solution says 768. They take 4!*2^5. Why is it not 5!?

Thanks

## Hi Harmon,

Hi Harmon,

This question type is actually a pet peeve of mine.

A lot of people (including some test prep companies) have posted similar questions, and the part I have a problem with is that they all assume that students recognize that some arrangements are considered IDENTICAL if the relative position of the people are the same.

Here's what I mean:

To simplify matters, let's say there are 4 (A, B, C and D) people seated in a circle, and that the four seats are: TOP, RIGHT, BOTTOM and LEFT.

So, one possible arrangement is: A-top, B-right, C-bottom and D-left

If each guest moves one chair to the left, the new arrangement is: D-top, A-right, B-bottom and C-left

If each guest moves one chair to the left (again), the new arrangement is: C-top, D-right, A-bottom and B-left

If each guest moves one chair to the left (again), the new arrangement is: B-top, C-right, D-bottom and A-left

All four of these arrangements are considered IDENTICAL since their relative positions (e.g., B sits to the A's left) are the same.

I don't believe I've ever seen an official GMAT question that tests this concept.

More importantly, IF the official test makers did create such a question, you can be guaranteed that they'd include some proviso that tells us arrangements are considered identical if the relative positions of the seated individuals are the same.

If we add that proviso regarding circular arrangements, then we can say:

The number of arrangements of n objects in a circle = (n-1)!

Cheers, Brent

## Hi Brent,

Thanks this helps. We can infer this only if its a circle and they give you the proviso right?

## That's correct.

That's correct.

## Hi Brent!

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

How do we know there are exactly half cases where X>Y?

## Question link: https:/

Question link: https://gmatclub.com/forum/runners-v-w-x-y-and-z-are-competing-in-the-ba...

Here's one way to explain it...

Let's start with 5 slots: _ _ _ _ _

Now place the letters V, W and Z in 3 of the slots.

So, for example, one possible outcome is: W_VZ_

At this point, if we place X and Y in the remaining two slots, there are two ways to accomplish this: WXVZY and WYVZX.

That is, in one case, X is ahead of Y, and in the other case Y is a head of X.

In fact, for every possible outcome, we can always interchange X and Y to get another outcome.

In other words, for half the possible outcomes, X is ahead of Y, and for the other half of the possible outcomes Y is ahead of X.

Another way to look at it is to examine a smaller case.

Let's say, for example, there are only 3 runners: W, X and Y

There are six possible outcomes: WXY, WYX, XWY, XYW, YWX, and YXW

In 3 of the outcomes X is ahead of Y, and in the other 3 outcomes Y is ahead of X.

Does that help?

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