Permutation Question.

Hi guys, first post here hope it is in the right category.

Here is the question from a set of questions I am currently working through.

"If no digit is to be repeated, how many 3 digit numbers can be formed with the digits; 0,1,2,3,4"

My thought was that you would simply be permeating the total 5 digits into 3 places, been 5!/2!=60 arrangements. However the textbook gives me an answer of 48 so I'm confused. This type of mathematics is not my strong point so hopefully someone can shed some light on were I am going wrong.

Thanks.

Jadon said:

Hi guys, first post here hope it is in the right category.
Here is the question from a set of questions I am currently working through.
"If no digit is to be repeated, how many 3 digit numbers can be formed with the digits; 0,1,2,3,4"
My thought was that you would simply be permeating the total 5 digits into 3 places, been 5!/2!=60 arrangements. However the textbook gives me an answer of 48

Click to expand...

60 would be correct if we allow \(\displaystyle 041\) to count as a three digit number. But it is a two digit number.
Therefore there only four choices for the hundredth digit: \(\displaystyle 4\times 4\times 3=48\)