There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
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There are: 9C6 = 84 combinations
This questions can be rewritten as 52 choose 6 or 52C6. This is the same as (52!)/(6!(52-6)!) (52!)(6!46!) (52*51*50*49*48*47)/(6*5*4*3*2*1) 14658134400/720 20358520 There are 20,358,520 combinations of 6 numbers in 52 numbers. This treats 1,2,3,4,5,6 and 6,5,4,3,2,1 as the same combination since they are the same set of numbers.
To calculate the number of 4-digit combinations you can get from the numbers 1, 2, 2, and 6, we need to consider that the number 2 is repeated. Therefore, the total number of combinations is calculated using the formula for permutations of a multiset, which is 4! / (2!1!1!) = 12. So, there are 12 unique 4-digit combinations that can be formed from the numbers 1, 2, 2, and 6.
You multiply all the numbers together. 6•3•4
(9 x 8 x 7 x 6 x 5 x 4)/(6 x 5 x 4 x 3 x 2) = 84 combinations.