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## Comment on

Product Codes## Why is i no 5x4x6x6?

5 for one of 5 digits

4 for one of 4 digits

6 for one of 6 letters

6 for 3!, the amount of possibilities of arrangements.

## Hi Philip,

Hi Philip,

The first two steps in your solution inadvertently count each outcome TWICE.

Here's what I mean: In step 1, you say there are 5 ways to select one of 5 odd digits, and in step 2, you say there are 4 ways to select one of remaining 4 odd digits. This means that the total number of ways to select two odd digits = 5 x 4 = 20.

To see where the mistake lies, consider the outcome where we choose the digit 9 in step 1, and we choose the digit 7 in step 2. This is one outcome.

HOWEVER, with your approach, this outcome is considered different from choosing digit 7 in step 1, and choosing digit 9 in step 2.

These two outcomes are the same.

So, to account for this duplication, we should take 20 and divide by 2 to get 10 outcomes.

Alternatively, we could just use combinations (5C2), as is used in the video.

## I'll add to my last post.

I'll add to my last post.

In your approach, the following 20 outcomes are considered different:

1, 3

3, 1

1, 5

5, 1

1, 7

7, 1

1, 9

9, 1

3, 5

5, 3

3, 7

7, 3

3, 9

9, 3

5, 7

7, 5

5, 9

9, 5

7, 9

9, 7

As you can see, each pair of outcomes are identical.

## How many 3 digit odd numbers

100s place can be accomplished using 3 ways

1s place can be accomplished using 2 ways ( since 0 is not allowed and one of 3 5 7 has been choose )

10s place can be accomplished using 2 ways ( since 2 number out of 0 3 5 7 has already choose )

answer will be 3*2*2 12 ways

but is the following answer correct if they have said no repetition

100s place can be accomplished using 3 ways

10s place can be accomplished using 4 ways

1s place can be accomplished using 3 ways.

therefore 36 ways if repetition allowed.

Please answer

## Yes, 36 is correct.

Yes, 36 is correct is repetition IS allowed.

## I have a doubt in this

100s place can be accomplished in 3 ways as 0 cant be the 100s place digit

10s place can be accomplished in 3 ways as repetition is not allowed. Ideally there are 4 ways but since one digit is used in 100s place so, 3 ways.

1s place can be accomplished in 2 ways since 2 digits are used in 100s and 10s place. so result is 3*3*2 which is 18. Please clarify.

## yogasuhas is saying that the

yogasuhas is saying that the answer is 36 IF the question were:

How many 3 digit odd numbers can be formed using 0,3,5,7 is repetition IS allowed?

The correct answer 36

If repetition is NOT allowed, then the correct answer is 12

In your solution, you are allowing for the possibility that the units digit is 0, which would make the number EVEN.

Cheers,

Brent

## Very confusing. I still don't

## REPEATED DIGITS ARE NOT

REPEATED DIGITS ARE NOT ALLOWED

Stage 1: Select hundreds digit

Since the hundreds digit cannot be 0, we can select 3, 5 or 7

So, we can complete stage 1 in 3 ways

Stage 2: Select units digit

The units digit must be odd. So, it can be 3, 5 or 7, HOWEVER we already selected an odd digit in stage 1.

Since repeated digits are NOT allowed, there are 2 odd digits to choose from.

So, we can complete stage 2 in 2 ways

Stage 3: Select tens digit

We have already selected 2 digits in stages 1 and 2.

Since repeated digits are NOT allowed, there are 2 remaining digits to choose from.

So, we can complete stage 3 in 2 ways

So, TOTAL number of outcomes = (3)(2)(2) = 12

------------------------------------------------

REPEATED DIGITS ARE ALLOWED

Stage 1: Select hundreds digit

Since the hundreds digit cannot be 0, we can select 3, 5 or 7

So, we can complete stage 1 in 3 ways

Stage 2: Select units digit

The units digit must be odd. So, it can be 3, 5 or 7.

Since repeated digits ARE allowed, there are 3 odd digits to choose from.

So, we can complete stage 2 in 3 ways

Stage 3: Select tens digit

Since repeated digits ARE allowed, the tens digit can be 0, 3, 5 or 7.

So, we can complete stage 3 in 4 ways

So, TOTAL number of outcomes = (3)(3)(4) = 36

Does that help?

Cheers,

Brent

## This is where I have a doubt,

## ZERO is, indeed, an even

ZERO is, indeed, an even integer.

EVEN integers: .....-6. -4, -2, 0, 2, 4, 6,.....

Cheers,

Brent

## if question asks for even

## If the question required us

If the question required us to create an EVEN integer, then stage 2 (selecting the units digit) can be completed in 1 way, since we MUST select the 0 to be units digit.

## John has 12 clients and he

## Here's my step-by-step

Here's my step-by-step solution: https://gmatclub.com/forum/john-has-12-clients-and-he-wants-to-use-color...

Cheers,

Brent

## Hi,

I don't understand why the answer is not just 5 X 4 X 6 = 120.

Since the digits cannot be repeated, shouldn't this use the Fundamental Counting principle in its simplest form?

Why are we doing this 3 times to get 360?

Thanks.

## When using the Fundamental

When using the Fundamental Counting Principle (FCP), you must be sure to clearly state/understand what is occurring at each stage (step).

You are saying that the 1st stage can be completed in 5 ways.

In your solution, what exactly is occurring during stage 1?

Are you selecting a letter or a number?

Likewise, what exactly is occurring during stages 2 and 3?

Once you state how you are setting up your solution, it will be much easier to identify where the problem lies.

Cheers,

Brent

## Hi,

Plz let me know if this approach will be valid

1st case: 1st digit odd nos, 2nd odd no. and 3rd digit letter i.e. = 5 * 4 * 6 = 120

2nd case: 1st digit letter, 2nd odd no. and 3rd digit odd no. i.e. = 6 * 5 * 4 = 120

3rd case : 1st digit odd nos, 2nd letter. and 3rd digit odd no. i.e. = 5 * 6 * 4 = 120

Therefore total no. of ways = 120 + 120 + 120 = 360

## That's a perfectly valid

That's a perfectly valid approach. Nice work!

## Crazy question...I highly

## It's a difficult question,

It's a difficult question, but I wouldn't say it's beyond the scope of the GRE.

That said, I'd say it's a 165+ level question, which means most people will not encounter a similar question on test day.

Cheers,

Brent

## for this questions couldnt we

## 5C2 * 6C1 = the total number

5C2 * 6C1 = the total number of ways to select 2 odd digits and 1 letter.

For example we might select 3, 7 and B

At this point we still haven't arranged the three characters.

We can take those 3 characters and arrange them in 3! different ways: B37, B73, 3B7, 7B3, 37B and 73B

So we must take 5C2 * 6C1 and multiply it by 3! (the number of ways to arrange the three selected characters)

Note: Although the question says the three characters can be arranged in any order (B73 or 3B7), it doesn't say that outcomes with the 3 same characters are considered the same.

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