Number of ways for a 6 digit license plate such that each digit is unique except for one The problem asks how many different six digit license plates can be made if there are three digits that occur once and one digit that occurs thrice?
My idea is that there can be 10 numbers for the reoccurring digits, and $6 \choose 3$ ways to lay them out. Then there are $9\cdot8\cdot7$ ways to lay out the remaining unique digits. This gives $10\cdot{6\choose3}\cdot9\cdot8\cdot7$ possible plates. 
Is this thought process correct? I'm having some trouble on these problems.
Thanks


Then there are $9\cdot8\cdot7$ ways to lay out the remaining unique digits. 

This is not about the number of ways of 'laying them out', but of picking digits for the three remaining digits.
So yes, you have indeed $6 \choose 3$ ways to lay out the recurring digit, and you have $10$ choices for what that digit is.
But then for the other three digits, you can pick $9$ choices for the 'first' (i.e. the first one you encounter from left to right in the digit string) digit, $8$ for the next, and $7$ for the last, giving you a total of:
$$10 \cdot {6 \choose 3} \cdot 9 \cdot 8 \cdot 7$$
Alternatively:
There are $9 \choose 3$ ways to pick the $3$ remaining digits, and $3!$ ways to lay them out:
$$10 \cdot {6 \choose 3} \cdot {9 \choose 3} \cdot 3!$$
which works out to be the same (of course!):
$$10 \cdot {6 \choose 3} \cdot {9 \choose 3} \cdot 3! =$$
$$ 10 \cdot {6 \choose 3} \cdot \frac{9!}{3!\cdot 6!} \cdot 3! =$$
$$10 \cdot {6 \choose 3} \cdot \frac{9!}{6!} =$$
$$10 \cdot {6 \choose 3} \cdot 9 \cdot 8 \cdot 7$$
