# Question: Product Codes

## Comment on Product Codes

### I understand that there are

I understand that there are three different ways to arrange DDL. But I can't wrap my mind around the fact that since the digits are actually different, that when you arrange three different object there are actually 6 different ways.
So what is the logic behind the fact that there are 3 different ways with this question? ### If you have 3 different

If you have 3 different objects to arrange, you can:
- Place the 1st object in one of 3 different locations (1st space, 2nd space and 3rd space)
- Place the 2nd object in one of the 2 remaining spaces
- Place the 3rd object in the 1 remaining space
So, the TOTAL # of arrangements = (3)(2)(1) = 6

Let's list them.
If there are three objects to arrange (A, B and C) we can create the following arrangements:
- ABC
- ACB
- BAC
- BCA
- CAB
- CBA

### So why do we just count

So why do we just count Digit1Digit2L, Digit1, L, Digit2 and L, Digit1, Digit2 as possibilities in this question, instead of the arrangements you just described? I thought your question about arranging 3 objects was separate from the question on the video.
I don't quite understand your latest question. Can you please rephrase it?

### Hi Brent,

Hi Brent,

In your second solution, we can solve using following way,

1. Select 2 Digit can also be solved as if order of two digit is important then it can be done = 5*4=20
2.Select One letter = 6
3. Arrange selected Letter in specific 2 digit by keeping order of digit important, this can be done in 3 ways, for ex if digit is 71 and letter is E then E can be placed as E71, 7E1, 71E. So a arrange letter in specific digit can be done =3

What you think? ### That's a perfectly valid

That's a perfectly valid approach - nice work!

That's what I love about GMAT math questions. They can often be solved in many different ways.

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

https://gmatclub.com/forum/fifteen-dots-are-evenly-spaced-on-the-circumference-of-a-circle-how-194198.html ### Hi Brent, Pardon me for

Hi Brent, Pardon me for saying this but it seems to me that the solution in the video is not right.
Here is what I am thinking. The three characters for the code can be selected in 5*4*6 = 120 ways. Now let's say this results in 37A. This code can be written in following 6 ways, i.e. 37A, 3A7, 73A, 7A3, A37, A73.
Hence the answer should be 120 *6 = 720 product codes are possible. Please can you clarify what am i missing?
Thanks. ### The only problem with your

The only problem with your solution is where you write: "The three characters for the code can be selected in 5*4*6 = 120 ways"
In your solution, 5*4 represents the number of ways to select two different odd digits.
The product 5*4 suggests that the order in which we select the two digits MATTERS when, in actuality, the order does NOT matter.
For example, in your solution, selecting the digit 3 first, and the digit 9 second is treated as DIFFERENT FROM selecting the digit 9 first, and the digit 3 second. However, those two outcomes are of the SAME.

Instead of using the Fundamental Counting Principle to select the two odd digits, we should be using combinations to select those two odd digits.
We can select 2 digits from 5 digits in 5C2 ways (= 10 ways)

Does that help?

### Thanks Brent. Got it. When I

Thanks Brent. Got it. When I have done 5*4 for selecting the 2 odd digits, I have accounted for the interchange of digits. i.e. 37 and 73 are both accounted for and these are 2 different cases included in the 20 ways. From here onwards, selecting a letter is possible in 6 ways and the letter can only be placed in 3 ways (not 3 factorial) with the 2 odd digits selected earlier (i.e. digit A can be placed with 37 in 3 ways - A37, 37A, 3A7). So 20*6*3=360. Thanks again.

### after selecting 5c2 , how to

after selecting 5c2 , how to again select 2 odd digits like we did in couple - committee problem ### Sorry, I'm not sure what you

Sorry, I'm not sure what you're asking.
I should point out, however, that there's a big difference between the above question and most committee questions.
And the above question, the order of the 2 digits and 1 letter matters.
For example, the code 17E is consider different from the code 7E1
In most committee questions, the order of the selected committee members doesn't matter.
For example, the committee ABC is the same as the committee BCA

But that help?