Hello,
How many even five-digit numbers are there?
The answer is 5*9*10^3
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
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 is number with a 3XXXX in the lead.
Then , eight choices for the lead digit, four choices for the 3 and choosing the three other digits.
If we add those together, .
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)