counting number of plates with no restrictions If there are no restrictions on where the digits and letters are placed, how many 8-character license plates consisting of 5 letters and 3 digits are possible if no repetitions of letters or digits are allowed?
my ans:
For the five letters there $26 \cdot 25\cdot 24 \cdot 23 \cdot 22 $ and for the digits $10 \cdot 9 \cdot 8$ so we have
$$ 26 \cdot 25\cdot 24 \cdot 23 \cdot 22 \cdot 10 \cdot 9 \cdot 8 $$
such a plates. But, my answer key says the answer is
$$ { 26 \choose 5} {10 \choose 3} \cdot 8! $$
I dont understand why this is the answer.
 A: I will break the key answer bit by bit, to try and give an intuition of why it is correct:


*

*${26\choose{5}}$ is the number of ways you have to pick $5$ letters from the alphabet without repetition;

*${10\choose{3}}$ is the number of ways you have to pick $3$ digits without repetitions.
At this point you have a bag with $5$ letters and $3$ digits. To assemble a plate, you need to order them in some way. There are $8!$ ways of ordering the contents of your bag, thus for each bag there are $8!$ plates. If there are ${26\choose{5}}{10\choose{3}}$ bags, then there are
$${26\choose{5}}{10\choose{3}}8!$$
plates.
Maybe we can also see why your answer doesn't quite work:
The product $26\cdot25\cdot24\cdot23\cdot22\cdot10\cdot9\cdot8$ is accounting for the number of ways you have to create a sequence of $5$ letters and $3$ numbers. You are not taking into account the fact that the numbers and letters can have their relative positions changed.

Take a smaller alphabet as a practical example: assume $A,B$ are the only letters and $1,2,3$ the only digits, and that a plate is a letter and two numbers. Your product was $2\cdot3\cdot2 = 12$, which would correspond to the following plates:
$A12, A13, A21, A23, A31, A32, B12, B13, B21, B23, B31, B32$
But there are many more plates; namely the ones obtained by permutating the contents of the above plates.
