# Thread: Counting Principles, five-digit number combinations

1. ## Counting Principles, five-digit number combinations

Hello,

How many even five-digit numbers are there?

I have 5 places _ _ _ _ _
I thought for a number to be even, the last place needs to contain an even number (0,2,4,6,8) which are 5 possibilities.

The answer indicates that some other place only can have 9 choices, instead of 10 (0-9). I do not understand why.

The following part of the question is

How many five-digit numbers are there with exactly one 3?
Here I do not even know where to start.

Thanks for any help

2. Originally Posted by DBA

The answer indicates that some other place only can have 9 choices, instead of 10 (0-9). I do not understand why.
If the first digit is 0, we do not consider it a 5 digit number any longer. (Usually stated as, disallow leading zeroes.)

Originally Posted by DBA
The following part of the question is

How many five-digit numbers are there with exactly one 3?
Here I do not even know where to start.

Thanks for any help
You can view it as one of 3xxxx, x3xxx, xx3xx, xxx3x, xxxx3 where x's are blanks (easier to type x).

3. Originally Posted by DBA
How many even five-digit numbers are there?
I have 5 places _ _ _ _ _

The answer indicates that some other place only can have 9 choices, instead of 10 (0-9). I do not understand why.

How many five-digit numbers are there with exactly one 3?
Here I do not even know where to start.
The reason for the 9 is your text does not allow leading zeros.
i.e 04542 does not count as a five-digit number.

For your second one $\displaystyle 9^4$ is number with a 3XXXX in the lead.
Then $\displaystyle 8\cdot 4\cdot 9^3$, eight choices for the lead digit, four choices for the 3 and choosing the three other digits.

If we add those together, $\displaystyle 9^3\cdot 41$.

4. I looked at every position:

1st position: All positions cannot contain a three
3XXXX 9*9*9*9=9^4

2nd position: First position cannot be zero and all positions cannot contain a three
X3XXX 8*9*9*9=8*9^3

3rd position:
XX3XX 8*9*9*9=8*9^3

4th position:
XXX3X 8*9*9*9=8*9^3

5th position:
XXXX3 8*9*9*9=8*9^3

They are disjoint events to me, so I would add them up:

9^4 + 8*9^3 + 8*9^3 + 8*9^3 + 8*9^3 = 9^4 + 4(8*9^3)

Thanks! The answer in the book says 9^4 + (4*8*9^3)