Cards are drawn from a deck of 52 with replacement. In how many ways can ten cards be drawn so that the tenth card is not a repetition?

Question

I am having trouble understanding the following problem:

Cards are drawn from a deck of $52$ with replacement. In how many ways can ten cards be drawn so that the tenth card is not a repetition?

I had initially thought the solution was:

$$52^9 \cdot (52-9)$$ ...as the first $9$ cards can be anything and the $10$th cannot be the same as the first $9$.

However, I was told that this was incorrect and the solution I was given by my professor was:

$$52 \cdot 51^9$$

Please confirm that the solution given to me was correct and/or explain how it was calculated.

Thank you.

Answer 1

The comment from @steven-gregory might be enough, but I'll elaborate a a little more. As @pbn990 pointed out, your original answer assumes that all the first 9 cards are different.

For simplicity, I'll write about the case of four cards:

Cards are drawn from a deck of 52 with replacement. In how many ways can four cards be drawn so that the fourth card is not a repetition?

Your original approach would give the answer $52^3 \cdot (52-3) $. But that assumes that all the first three cards are the same. For example, assume these are the cards drawn:
Jack of Spades, Jack of Hearts, Jack of Spades, (??)

For the fourth card, there are NOT $\; "52-3"$ options (cards) that fulfill the question's requirements, but instead there are $\; "52-2"$ options. You would have to subtract only two, because there are only two cards that would make the fourth a repetition: Jack of Spades and Jack of Hearts.

Now, the recommended solution implies starting counting (the number of ways) from the fourth card. You may wonder why you can start counting from the fourth card. That's because of two reasons:

1. Cards are drawn with repetition
2. Each card extraction is independent from all the others

Reason 2 is a consequence of reason 1.
For the calculation, the fourth card can be any card, so there are 52 ways of choosing it. And we know that the fourth card is not a repetition, so the third card cannot be the same as the fourth. Therefore, there are only $\; "52-1"$ cards to choose from. For the second card, it doesn't matter if it's the same as the third card. We only want the second card to be different from the fourth card, thus leaving us with $\; "52-1"$ cards that we can choose from if we want to fulfill the requirement. It's the same for the first card. And that gives us the final answer of $$ 51^3 \cdot 52 $$