probability of throwing 5 different numbers on 6 dices

We throw 6 dices, what is the probability of 5 different numbers.

I do not understand the solution:

$$P(5 \, \text{different}) = \frac{\binom{6}{2}\cdot 6 \cdot 1 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{6^6}$$

I understand the sample space : $6^6$,

however I do not understand the numerator if somebody can explain what it means.

Is there a more intuitive way to solve this problem?

Solution 1:

You noted that you already understand the total number of possible outcomes, so it only remains to determine the number of outcomes with 5 different numbers.

If six dice roll 5 different values, it must be the case that one number is rolled twice and the others are rolled once. There are $\binom{6}{2}$ ways to choose which dice are the double, and 6 numbers the diuble could be. Then there are four numbers which are rolled once, for which there are $\binom{5}{4}$ choices, and $4!$ ways to assign them to the 4 singleton dice.

Then the number of possibilities with 5 distinct rolls is:

$$N=\binom{6}{2}\cdot6\cdot\binom{5}{4}\cdot4!\\ =\binom{6}{2}6!$$

Which turns out to be the numerator in the probability calculation.