# Question: 3-Digit Odd Numbers

## Comment on 3-Digit Odd Numbers

### I thought 3 digit odd numbers

I thought 3 digit odd numbers means all the numbers must 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.

### 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,

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

In stage 2, I multiplied 5by2 because the place of the digit is not determined. So I have 5 numbers to choose from and 2 places for it , 5*2=10

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

My task is choose a number and put it in one of the 2 places remaining!

Thanks ### 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

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

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

I don't understand why, in the video example above, you reason that there are 5 digits to choose from in the hundred digit slot. The unit digit slot can only contain 2 numbers (which are odd).

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

Does this question means that no digit can be repeated in the 3 digit no like 123, 456, .... What after that , question says, digits cannot be repeated , so does that means digit 1 cannot be repeated in second 3 digit no ?? ### 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