If there are no restrictions on where the digits and letters are placed, how many $8$-place license plates consisting of $5$ letters and $3$ digits are possible if no repetitions of letters or digits are allowed? What
if the $3$ digits must be consecutive?
The answer to the first part is $10 \choose 3$$26 \choose 5$$*8!$
The answer to the second part is $10 \choose 3$$26 \choose 5$$*6!*3!$
My thought process for the first part is we choose three numbers and five letters, then for the first place there are eight ways to order them, then seven, six, etc.
For the second part, I would think the two factorials would have been $5!*3!$ not six and three.
Are there any errors in my reasoning for the first answer? Why are the factorials in the second part $6!*3!$?
Best Answer
You are correct for the first problem, though I would rather think of $8!$ as a single high-level arranging operation rather than eight separate placements of the characters. I view this enumeration as consisting of three steps.
Step 1: Choose three distinct numbers in $\binom{10}{3}$ ways
Step 2: Choose five distinct letters in $\binom{26}{5}$ ways
Step 3: Arrange these eight characters in $8!$ ways
You can approach the second problem similarly if you think of the digits as comprising a single, indivisible unit (though the digits may be mixed up among themselves).