Question: Product Codes

Comment on Product Codes

Why is i no 5x4x6x6?

Why is it not 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

How many 3 digit odd numbers can be formed using 0,3,5,7, repetition not allowed.

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. Yes, 36 is correct.

Yes, 36 is correct is repetition IS allowed.

I have a doubt in this

I have a doubt in this comments question. If we have the restriction of repetition not allowed, then how can 36 be the answer? I mean-

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?

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

Very confusing. I still don't understand. How would you take the digits? please explain. 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,

This is where I have a doubt, in stage 3(where repetition is not allowed)how do we have 2 digits to choose from? Can 0 be chosen in the tens place as 0 is considered an even number? Thank you so much for explaining :) 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 digit rather than odd do we need to consider 0 for the second stage? 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

John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes. how to solve Hi,

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,

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!