The full house term (3+2) I agree with. For 2+2+1, you have a choice of 3 digit values for the pairs (3C2), but the single digit can then be any digit value not chosen: 5C1. Similarly in the 2+1+1+1 term.joshmccraney said:so something like this (for now i'll ignore the zero): [tex]\underbrace{{2\choose1}\frac{5!}{2!3!}}_{\text{both full houses}}+\underbrace{{3\choose2}{3 \choose 1}\frac{5!}{2!2!}}_{\text{3 of the two pairs without a lone 6}}+\underbrace{{3\choose1}\frac{5!}{2!}}_{\text{three of the single pairs}}+\underbrace{{6\choose5}5!}_{\text{no pairs}}[/tex] is this correct (or at least close)? my choose notation is for selecting which of what to use and the factorials are for counting the arrangements once we have chosen what to use. again, I've disregarded zero for now, but I'm not sure this is complete even without zero. any help please!
To calculate the total number of unique 5-digit numbers from a given list, you need to use the formula nPr = n!/(n-r)!, where n is the number of digits in the list and r is the number of digits in the number. This will give you the total number of permutations, which is the total number of unique 5-digit numbers.
Yes, you can use the same digit more than once when calculating unique 5-digit numbers. In fact, using the same digit multiple times will increase the number of possible combinations and ultimately result in a larger number of unique 5-digit numbers.
To determine the probability of getting a specific 5-digit number from a given list, you need to first calculate the total number of unique 5-digit numbers from the list using the nPr formula. Then, divide the number of times the specific number appears in the list by the total number of unique 5-digit numbers. This will give you the probability of getting that specific number.
Yes, there is a limit to the number of unique 5-digit numbers that can be calculated from a given list. The maximum number of unique 5-digit numbers that can be calculated from a list of n digits is n^5. This limit is reached when all possible combinations of the n digits are used to create the 5-digit numbers.
Yes, you can use this method to calculate unique numbers with more or less than 5 digits. The nPr formula can be adjusted to accommodate different numbers of digits. For example, if you want to calculate unique 3-digit numbers from a given list, you would use the formula nPr = n!/(n-r)! where n is the number of digits in the list and r is the number of digits in the number (in this case, 3).