Probability 5 dice have distinct numbers

Poker dice is played by simultaneously rolling 5 dice. What is the probability all five dice have distinct numbers?

My work:

As the dice are the same/ indistinguishable, order doesn't matter (that is if die 1 rolls 1, die 2 rolls 2, die 3 rolls 3 and die 4 rolls 4 and die 5 rolls 5 is the same as die 1 is 2, die 2 is 1, die 3 is 3 and die 4 is 4 and die 5 is 5)

There are 6*5 = 30 possible numbers to be chosen (5 number 1, 5 number 2 ... and so on to 5 number 3)
and since we roll 5 dice, the number of possible outcomes is 30 choose 5

There are (6 choose 5) ways to choose the number of the die to be distinct (out of 1,2,3,4,5,6) and (5 choose 1) ways to choose that number (ie; if you choose 1, there are 5 number 1s)

So my answer is (6 choose 5)*(5 choose 1)^5 / (30 choose 5) = 0.13157

The answer is actually 6*5*4*3*2/6^5 = 0.09259. That makes sense but only if the dice are distinct. How do you solve this problem using combinations?

Baron said:

Poker dice is played by simultaneously rolling 5 dice. What is the probability all five dice have distinct numbers?

Click to expand...

If we toss a die five times then there are \(\displaystyle 6^5\) possible outcomes.

There are \(\displaystyle _6\mathcal{P}_5\) ways to have all five dice different.

pka said:

If we toss a die five times then there are \(\displaystyle 6^5\) possible outcomes.

There are \(\displaystyle _6\mathcal{P}_5\) ways to have all five dice different.

Click to expand...

I understand the solution but I'm confused as to what was wrong with my reasoning using combinations.