Have questions about your preparation or an upcoming test? Need help modifying the Study Plan to meet your unique needs? No problem. Just book a Skype meeting with Brent to discuss these and any other questions you may have.

- Video Course
- Video Course Overview
- General GMAT Strategies - 7 videos (free)
- Data Sufficiency - 16 videos (free)
- Arithmetic - 38 videos
- Powers and Roots - 36 videos
- Algebra and Equation Solving - 73 videos
- Word Problems - 48 videos
- Geometry - 42 videos
- Integer Properties - 38 videos
- Statistics - 20 videos
- Counting - 27 videos
- Probability - 23 videos
- Analytical Writing Assessment - 5 videos (free)
- Reading Comprehension - 10 videos (free)
- Critical Reasoning - 38 videos
- Sentence Correction - 70 videos
- Integrated Reasoning - 17 videos

- Study Guide
- Blog
- Philosophy
- Office Hours
- Extras
- Prices

## Comment on

3-Digit Odd Numbers## I thought 3 digit odd numbers

## Not quite.

Not quite.

ODDS: ...-7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, etc.

More here: https://www.gmatprepnow.com/module/gmat-integer-properties/video/837

## I thought this was a

I thought this was a combination question that could be solved using combination formula, but you didn't. Won't it be faster using the formula?

## This lesson appears at the

This lesson appears at the beginning of the Counting module, before we cover combinations. So, the solution uses the Fundamental Counting Principle only.

How about showing us how you'd use combinations to solve the question.

ASIDE: I'm not sure how much faster the solution will be if we use combinations.

## Hi Brent,

After we selected the 3rd most restrictive digit, we have to select the 2nd or 1st digit. We are left with 5 numbers and 2 places so why we did not multiply 5*2? and for the last digit 4*1. So the answer is 80. What am I doing wrong in here?

## When solving counting

When solving counting questions it helps to identify EXACTLY what is accomplished during each stage:

Stage 1: Choose the 3rd digit

Stage 2: Choose the 1st digit

Stage 3: Choose the 2nd digit

Stage 1: There are 2 digits to choose from. So, we can complete stage 1 in 2 ways

Stage 2: At this point, there are 5 digits remaining. So, we can complete stage 2 in 5 ways. (not 10 ways as you suggest).

NOTE: Keep in mind that our sole goal in stage 2 is to choose a single digit to go in the hundreds place.

Stage 3: At this point, there are 4 digits remaining. So, we can complete stage 3 in 4 ways.

Does that help?

Cheers,

Brent

## In stage 2, I multiplied 5by2

For the last number I have only 4 numbers with just one place for it so 4*1=4

## What is your task in stage 2?

What is your task in stage 2?

Mine is: "Choose the 1st digit" (in other words, "Choose a digit to go in the hundreds position")

Cheers,

Brent

## My task is choose a number

Thanks

Mohammad

## I see! Unfortunately, that

I see! Unfortunately, that setup will cause duplication.

For example, in your stage 1, we might choose 3 for the units digit.

In your stage 2, we might choose 6 for the HUNDREDS digit.

In your stage 3, we might choose 8 for the TENS digit.

So, 683 would count as 1 possible outcome.

Here's where the duplication comes in.

In your stage 1, we might choose 3 for the units digit.

In your stage 2, we might choose 8 for the TENS digit.

In your stage 3, we might choose 6 for the HUNDREDS digit.

So, 683 would count as another possible outcome.

HOWEVER, these two outcomes are the same.

In fact, your method counts ever outcome twice.

So, to account for this duplication, we must take your answer of (2)(5)(2)(4)(1) and divide it by 2 to get 40

To avoid this problem, stay away from creating 2-step tasks like "Choose a number AND place it in 1 of 2 places"

I hope that helps.

Cheers,

Brent

## Yes it did.

Thanks

Mohammad

## why didn't we do 5C2 here?

filling 2 places with 5 numbers. Why is 5c2 wrong here?

## Since the order of the two

Since the order of the two selected digits matters, we cannot use combinations.

Let's say the 5 digits are 1, 2, 3, 4, 5

5C2 = 10

So, there are 10 ways to select 2 digits (if order does not matter):

1,2

1,3

1,4

1,5

2,3

2,4

2,5

3,4

3,5

4,5

However, if we agree that 12 is different from 21 (i.e., order matters), then there are 20 different arrangements:

12

13

14

15

21

23

24

25

31

32

34

35

41

42

43

45

51

52

53

54

Does that help?

Cheers,

Brent

## I don't understand why, in

So if two numbers can only be occupied in the unit digit slot, and if each number cannot be repeated then the numbers that are used in the unit digit slot (1 and 5) cannot be used in our hundred digit slot). Therefore, only 4 of the remaining digits can make up the hundred digit slot. And 3 of the remaining digits in the tens digit slot. Right?

I ended up getting 48: Unit digit slot = 2

Tens digit slot = 3

Hundred digit slot = 4

2 x 3 x 4 = 24

Since the digits in the tens and hundreds places can be reversed to create a different number (i.e. 125 or 215), I multiplied 24 x 2 = 48

What am I doing wrong?

## It is incorrect to say

It is incorrect to say "Therefore, only 4 of the remaining digits can make up the hundred digit slot."

For example, if we choose a 1 for the units digit, then we can select 2, 4, 5, 6, or 8 for the hundreds digit (5 choices).

If we choose a 5 for the units digit, then we can select 1, 2, 4, 6, or 8 for the hundreds digit (5 choices).

There's nothing in the wording of the question that says the units digit and the hundreds digit cannot both be odd.

So, some possible outcomes include: 145, 561, 581, 125, etc

Does that help?

Cheers,

Brent

## Does this question means that

## The question says that EACH

The question says that EACH number cannot contain repeated digits.

So, within a single 3-digit number, we cannot have more that 1 of each digit.

So, for example, 861 is fine, but 811 is not fine.

Does that help?

Cheers,

Brent

## I suddenly understand why we

## Exactly!

Exactly!