Solving Odd Digit Divisibility by Five using Permutations

[hms.tech]

Homework Statement

How many four-digit numbers formed of only odd digits are divisible by five?

Homework Equations

Permutations

The Attempt at a Solution

Here is what I think should be done :

Ans : 4P3 * 1
= 24

Is that right ?

 

[HallsofIvy]

You don't say how you got that number so I don't see any way to comment except to say that 4P3= 4 is clearly NOT the correct answer. You don't say why you think that is true. How did you get that?

There are a total of 4 digits in the number and 5 odd digits. How many choices are there for the first digit? The second ? The third? The fourth?

 

[tiny-tim]

Ans : 4P3 * 1
= 24

why?

 

[hms.tech]

You don't say how you got that number so I don't see any way to comment except to say that 4P3= 4 is clearly NOT the correct answer. You don't say why you think that is true. How did you get that?

There are a total of 4 digits in the number and 5 odd digits. How many choices are there for the first digit? The second ? The third? The fourth?

here is how I did it :

The last digit is reserved for "5" since we want it to be divisible by "5"

Then, the choices for the first digit are : 4
2nd digit : 3
3rd digit : 2
Ergo, 4P3 * 1 = 4P3 = 24

I am honestly surprised why this method is incorrect .

 

[tiny-tim]

ah, you're assuming they all have to be different

they don't ​

(btw, I'm not familiar with this 4P3 notation, but it doesn't look right …

24 = 4!, so where does 3 come into it? )

 

[hms.tech]

ah, you're assuming they all have to be different

they don't ​

(btw, I'm not familiar with this 4P3 notation, but it doesn't look right …

24 = 4!, so where does 3 come into it? )

Hmmm...I think you are right, I must not assume this.

Honestly, I have no experience how to calculate the permutations if digits can be repeated.

I'll try anyway :

There are 5 choices for each of the first three digits and one choice for the last digit.
5*5*5*1 = 125

 

[Dick]

Hmmm...I think you are right, I must not assume this.

Honestly, I have no experience how to calculate the permutations if digits can be repeated.

I'll try anyway :

There are 5 choices for each of the first three digits and one choice for the last digit.
5*5*5*1 = 125

Right!

 

[tiny-tim]

Woohoo! ​

 

[HallsofIvy]

Hmmm...I think you are right, I must not assume this.

Honestly, I have no experience how to calculate the permutations if digits can be repeated.

If the digits can be repeated, this is NOT a permutations problem. It is simply a application of the "fundamental counting principle": if A can be done in m ways and B can be done, independently, in n ways the A and B can be done in mn ways.
There are 5 ways to choose the first digit, 5 ways to choose the second digit, 5 ways to choose the third digit, and only one way to choose the last digit which u must be 5.
5(5)(5)(1)=

I'll try anyway :

There are 5 choices for each of the first three digits and one choice for the last digit.
5*5*5*1 = 125

Exactly right.