Hello,
Make all possible number using the digits 1 , 3 , 7?
First of all I want to know how many number will be there and then the method to form the number?
Thanks.
Assuming those digits cannot repeat, there are $\displaystyle 3!=1\times2\times3= 6$ different ways to arrange those digits. Hence there are 6 different numbers that can be made.
EDIT
Another way of looking at it is that say I have those digits 1, 3, and 7.
Then I could place them in three spots _ _ _.
Choose a digit, lets say the 1. I can place that 1 in any of those three spots, e.g. _ _ 1. So far we have $\displaystyle 3\times ?\times ?$.
Now choose another digit, lets say the 7. There are only two spots where the 7 could be placed, so we have $\displaystyle 3 \times 2 \times ?$. Lets say we place it like so, 7 _ 1.
Now there is only one spot where the digit 3 can go so we have $\displaystyle 3 \times 2 \times 1 = 6$. Thus there are 6 different possible numbers.