The answer is 10!/[6!*(10-6)!] where n! represents 1*2*3*...*n
Number of combinations = 10*9*8*7*6*5*4*3*2*1/(6*5*4*3*2*1*4*3*2*1)
= 10*9*8*7/(4*3*2*1) = 210
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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.
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
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.
If you are assuming you can repeat any of the numbers then: nx where n=number of options and x= number of spaces using the formula assuming all numbers 0-9 then: 1010 10,000,000,000 is your answer. If numbers can not repeat and have to be used we use the formula: n! Where n is the number of options.Since you want 10 numbers we will do 10! 10! 10*9*8*7*6*5*4*3*2*1 362880 is your answer.