I assume you mean "six numbers" rather than "sox numbers". If the numbers are all distinct (i.e none of them are in the set of thirty numbers more than once), then there are 30!/(24!6!) ways of choosing six numbers, where "!" is the factorial of that number.
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?"