1000
90. all the numbers from 10 through 99
four, five, nine :)....and zero
The number line from 1 to 10 includes 10 integers, but an infinite number of real numbers.
1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10
The answer is 20C10 which is 20!/[10!(20-10)!] = 184756
When you spell out the words for numbers 1-10 four of the numbers have three letters. (one two six ten)
1000
90. all the numbers from 10 through 99
four, five, nine :)....and zero
3. Three, Seven, Eight.
Only the number 8 (VIII)
3 This are Three, Seven, and Eight.
The number of combinations of 7 numbers from 10 is 10C7 = 10*9*8/(3*2*1) = 120
109
Infinitely many. In fact there are more irrational numbers between 1 and 10 than there are rational numbers - in total!
There are [9 x 10^2001] numbers which have 2002 digits. The [^] symbol means raise to the power. So 10^2 is 10² = 100 and 10^3 is 10³ = 1000, etc.To see how this works, consider how many 2 digit numbers there are. The largest 2-digit number is 99 and there are 100 numbers [0 through 99] up to that. But we need to subtract off the 1-digit numbers [0 through 9]. There are ten of those, so we have 100 - 10 = 90.Or write it this way: 10^2 - 10^1 = 9 x 10^1. So for a number of digits (n) we have 9 x 10^(n-1).Let's check with n=3: 9 x 10^2 = 9 x 100 = 900. For 3-digit numbers, there is 100 through 999, so 0 through 999 is 1000 numbers [10^3], and subtract off 0 through 99 [100 or 10^2] which gives us 900 numbers, and satisfies the formula.