The number of combinations of 35 things taken 5 at a time is 35! / (5! * (35-5)! which is 35! / (5! * 30!). 35! / 30! is 35 * 34 * 33 * 32 * 31, and 5! is 120. After cancelling we are left with 7 * 34 * 11 * 4 * 31 which is 324,632
To find the number of different combinations of the numbers 1 to 10, we can consider the combinations of choosing any subset of these numbers. The total number of combinations for a set of ( n ) elements is given by ( 2^n ) (including the empty set). For the numbers 1 to 10, ( n = 10 ), so the total number of combinations is ( 2^{10} = 1024 ). This includes all subsets, from the empty set to the full set of numbers.
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
35
9000
None. You do not have enough numbers to make even one combination.
Assuming that the six numbers are different, the answer is 15.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
Their is 25 combinations
Only one.
1000
35
9000
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
None. You do not have enough numbers to make even one combination.
Oh, dude, you're making me do math now? Alright, so if you have three numbers and you're asking how many combinations you can make with those three numbers, it's like a little math puzzle. Each number can be used multiple times, so it's like a little party for those numbers. The total number of combinations you can make with three numbers is 27. That's like having 27 different outfits to choose from for a night out, but with numbers.
To calculate the number of combinations with three numbers, you would use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n is the total number of numbers you have to choose from, and r is the number of numbers you are choosing. So, if you have three numbers to choose from, there would be 3C3 = 3! / 3!(3-3)! = 6 / (6*0!) = 6 / 6 = 1 combination.