1. ## make number

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.

2. Assuming those digits cannot repeat, there are $3!=1\times2\times3= 6$ 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 $3\times ?\times ?$.

Now choose another digit, lets say the 7. There are only two spots where the 7 could be placed, so we have $3 \times 2 \times ?$. Lets say we place it like so, 7 _ 1.

Now there is only one spot where the digit 3 can go so we have $3 \times 2 \times 1 = 6$. Thus there are 6 different possible numbers.

3. ## Duh !!!

Take them to be a,b and c.
Now the diffferent possibilities are -
1. abc
2. acb
3. bac
4. bca
5. cab
6. cba

4. Originally Posted by hanuman_2000
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.
here three numbers given,6 different numbers can be written
that is 137,173,317,371,731,713
as 1 is the first digit - 2 numbers,
as 3 is the first digit - 2 numbers,
as 7 is the first digit- 2 numbers,
totally 6 numbers can be written.