This is a question about license plates that for the life of me I can't figure out why it's wrong. The question is: A license plate have 3 letters and 3 numbers. What are the total number of possible license plates if there is no restriction on which order letters and numbers appear? My answer was:
$10^3 \cdot 26^3 \cdot 6!$. The answer however says that in place of 6! it's
$10^3\cdot 26^3 \cdot \binom 6 3$ instead. I can't understand why it's not $6!$, my reasoning is that I treat each number and letter as its own entity and they can be arranged in $6!$ ways?
Best Answer
You choose which of the three positions will have letters. There are $\binom 6 3 = 20$ ways to do that. Then you choose the three digits and the three letters. There are $26^3 10^3$ ways to do that.
If you first choose three digits and three letters and then decide to shuffle them, it might seem plausible that there are $10^3\cdot26^3\cdot6!$ ways to do that. But suppose you choose $ABB262$. There are $6!=720$ orders in which to place six distinguishable objects. But these are not distinguishable: suppose you interchange the two $2$s and also the two $B$s: you get a different item in the list of $720$, but it should be counted as the same item in the list that you want. There are many similar cases, so you get a substantial overcount.