(45 x 44 x 43 x 42 x 41 x 40 x 39)/(7 x 6 x 5 x 4 x 3 x 2) = 45,379,620 of them.
Could you generate a complete set of 6 number combinations from 45 numbers ?
There are an infinite number of combinations where this could be the answer.
By making a number tree that could have as many as 1,000,000 combos.
Most credit cards use 16 digit numbers. American Express uses 15. Either one of them allows for so very many more combinations than there are, or could reasonably be, active credit card accounts that a number chosen at random is unlikely to belong to an actual real account.
There are 21 combinations.
Could you generate a complete set of 6 number combinations from 45 numbers ?
There are an infinite number of combinations where this could be the answer.
By making a number tree that could have as many as 1,000,000 combos.
Most credit cards use 16 digit numbers. American Express uses 15. Either one of them allows for so very many more combinations than there are, or could reasonably be, active credit card accounts that a number chosen at random is unlikely to belong to an actual real account.
There are 21 combinations.
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
Infinite. If numbers can be repeated, the list could go on nonstop.
You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.
6^4 = 1296 combinations but some are repeatable e.g. 1221 = 2121 = 2112 etc. so for the total number of non repeatable combinations with 4 dice, use pascals triangle to get 126 unique combinations.
There is no single answer to that. A lot of combinations of 6 numbers could have a mean of 18.5, as any 6 numbers that add to 111 could.
36
There are five vowels. Notice that any of the vowels could be chosen from 0 though 5 times in any given combination. We therefore need the number of combinations of 5 from 5 _with replacement_. This is the number ( 5 + 5 - 1 ) choose 5 = 126.