The number of combinations is 50C6 = 50*49*48*47*46*45/(6*5*4*3*2*1) = 15,890,700
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.
you could make a probability tree if you could be bothered
35
9000
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.
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.
you could make a probability tree if you could be bothered
Only one.
Assuming that the six numbers are different, the answer is 15.
35
9000
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
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.
None. You do not have enough numbers to make even one combination.
7
You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.