How many different four-digit numbers can be formed from the digits 0 through 9 if the first digit must be even and cannot be zero?
Ok So I can only use 2 4 6 8 so Is my answer 4000 or is it
3996? because it is 2000 - 2999 4000 - 4999 6000 - 6999 and 8000- 8999.
2007-02-11 04:57:51 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I think it is 4000. Why?
Like you said, you only have 4 possibilities for the thousands-digit and 10 for the others.
4*10*10*10 = 4000.
Actually, the long way you did it above is correct, except that when you consider 2000-2999 to be possibilities, the number of numbers is not 999, but 1000. You have to count the 2000 as 1 possibility.
For example, How many numbers are between 0 and 2 (if we include 0 and 2)
we have 0, 1, and 2----3 numbers
Not 2-0 = 2!
So, yes.. it is 4000
2007-02-11 05:06:27 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Both methods give you 4000, because 2000-2999 is 1000 numbers not 999 numbers.
2007-02-11 05:04:04 · answer #2 · answered by rscanner 6 · 0⤊ 0⤋
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2016-12-04 01:15:27 · answer #3 · answered by molander 3 · 0⤊ 0⤋
4 choices for the first digit. 10 choices each for the other 3. No dependencies among the choices.
4*10*10*10 = 4000
It's as simple as that.
Stop worrying; you're doing fine!
2007-02-11 05:06:42 · answer #4 · answered by Curt Monash 7 · 0⤊ 0⤋
Sorry Iam not a computer.
2007-02-11 05:00:17 · answer #5 · answered by Anonymous · 0⤊ 0⤋