Assuming no repetition is allowed, there are 8582777280 ways in which you can pick any of the numbers from 1 to 99 inclusive with 5 numbers. This is given by the formula n! / (n - r)! where n is the number of numbers you have to choose from, and r is the number of numbers chosen.
With whole numbers, there are five ways.
20C2 = 190
There is an infinite number of ways three numbers can have the sum of 11
When you have a choice of 7 digits, and cannot repeat any digit, then the first number can be chosen in 7 ways, the second can be chosen in 6 and the third can be chosen in 5. 7 x 6 x 5 = 210
several ways
The first (hundreds) digit can be chosen in any one of 9 ways from the digits {1,2,3,...,9}. With each one of these, the second (tens) digit can be chosen in any one of 10 ways from the digits {0,1,2,3,...,9}. With each of these combinationsm the third (units) digit can be chosen in one of 5 ways from the digits {1,3,5,7,9}. Altogether, then, there are 9*10*5 = 450 ways or 450 odd 3-digit numbers.
4
Assuming no repetition is allowed, there are 8582777280 ways in which you can pick any of the numbers from 1 to 99 inclusive with 5 numbers. This is given by the formula n! / (n - r)! where n is the number of numbers you have to choose from, and r is the number of numbers chosen.
With whole numbers, there are five ways.
10
One... numbers
three
There are 11880 ways.
20C2 = 190
24 ways
*facepalms* "Can you please repeat the question?"