The problem involves calculating the number of unique 5-digit numbers that can be formed from a given list of digits, which includes repetitions and the constraint that zero cannot be the leading digit.
The discussion is active, with participants offering various approaches to tackle the problem. Some suggest counting methods based on different digit combinations and arrangements, while others express uncertainty about the completeness of their reasoning, particularly regarding the inclusion of zero.
There is an assumption that zero cannot be the first digit, and participants are exploring how this affects the overall counting of unique numbers. The problem involves a specific list of digits with repetitions, which adds complexity to the calculations.
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): \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}} 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!