( 36! )/( 32! ) ( 4! ) = (36 x 35 x 34 x 33)/(4 x 3 x 2) = 58,905
two
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.
Assuming there is no repetition, you want the number of groups of 4 numbers, and the order doesn't matter: this is calculated as 33 over 4; which is the same as (33 x 32 x 31 x 30) / (1 x 2 x 3 x 4)
4!=4x3x2x1=24
To calculate the number of combinations of four numbers from a set of 32 numbers, you can use the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 32 (total numbers) and r = 4 (numbers chosen). Plugging these values into the formula, you get 32C4 = 32! / (4!(32-4)!) = 32! / (4!28!). After calculating this expression, you will find that there are 35,960 possible combinations of four numbers from a set of 32 numbers.
There are 35C4 = 35*34*33*32/(4*3*2*1) = 52,360 combinations.
10 * * * * * That is just plain wrong! It depends on how many numbers in each combination but there are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations.
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
( 36! )/( 32! ) ( 4! ) = (36 x 35 x 34 x 33)/(4 x 3 x 2) = 58,905
5*4*3*2*1 = 120 combinations. * * * * * No. The previous answerer has confused permutations and combinations. There are only 25 = 32 combinations including the null combinations. There is 1 combination of all 5 numbers There are 5 combinations of 4 numbers out of 5 There are 10 combinations of 3 numbers out of 5 There are 10 combinations of 2 numbers out of 5 There are 5 combinations of 1 numbers out of 5 There is 1 combination of no 5 numbers 32 in all, or 31 if you want to disallow the null combination.
16
Oh, isn't that a happy little question? Let's see here... To find the number of combinations of 4 numbers out of 7, we can use a simple formula: 7 choose 4, which is calculated as 7! / (4! * (7-4)!). So, there are 35 unique combinations of 4 numbers that can be made from a set of 7 numbers. Isn't that just delightful?
You would get 4!/2! = 12 combinations.
8C4 = 70
two
The answer is 32!/(27! * 5!) where n! represents 1*2*3*... *n So the answer is 32*31*30*29*28/(5*4*3*2*1) = 201 376