Solving dice problem using the inclusion-exclusion principle

Question

We roll a die ten times. What's the probability of getting all the six different numbers of the dice?

If $A_i$ is the event of getting at least of the numbers from $i=1$ to $6$. What the problem is asking is $P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \cap A_6)$ . So I guess I will get this probability if I develop the exclusion inclusion formula for the union of different events. Is the starting point to find $P(A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6 \cup A_7 \cup A_8 \cup A_9 \cup A_{10})$ ? Where $A_i$ would be here each throw of the dice?

We were asked to do it by the inclusion exclusion principle,as the tittle of the problem says. TIA for any hint or answer.

Answer 1

If there were no restrictions, there would be six possible outcomes for each of the ten throws, so there are $6^{10}$ possible sequences of throws. From these, we must exclude those in which fewer than six outcomes occur.

There are $\binom{6}{k}$ ways to exclude $k$ of the $6$ outcomes and $(6 - k)^{10}$ possible sequences involving only the remaining $6 - k$ possible outcomes of a die throw. By the Inclusion-Exclusion Principle, the number of ways all six outcomes can occur when a six-sided die is tossed ten times is $$\sum_{k = 0}^{6} (-1)^k\binom{6}{k}(6 - k)^{10} = \binom{6}{0}6^{10} - \binom{6}{1}5^{10} + \binom{6}{2}4^{10} - \binom{6}{3}3^{10} + \binom{6}{4}2^{10} - \binom{6}{5}1^{10} + \binom{6}{6}0^{10}$$

Answer 2

The problem can also be solved using classification. Let $n_1,n_2,\ldots,n_6$ be the number of dices showing $1, 2,\ldots,6$, respectively. Define the set

$$A_{n,r}=\{(n_1,\ldots,n_6)_{\,}|_{\,}n_1\geq r,\cdots,n_6\geq r, n_1+\cdots+n_6=n\}.$$

Then the number of ways to let all numbers $1,2,\ldots, 6$ appear at least once in $10$ dice throws is

$$S=\sum_{(n_1,\ldots,n_6)\in A_{10,1}}\frac{10!}{n_1!\cdots n_6!}.$$

Use this number to divide by $6^{10}$, you get the probability. There are ${9\choose 5}=126$ elements in $A_{10,1}$. We don't need to enumerate them one by one. Many are equivalent up to permulations. So one can divide them into $5$ classes:

$$\begin{array}{|c|c|c|} \hline \mbox{No.} & \mbox{Class} & \mbox{Count} & n_1!\cdots n_6!\\ \hline 1 & 5, 1, 1, 1, 1, 1 & 6 & 120\\ 2 & 4, 2, 1, 1, 1, 1 & 30 & 48\\ 3 & 3, 3, 1, 1, 1, 1 & 15 & 36\\ 4 & 3, 2, 2, 1, 1, 1 & 60 & 24\\ 5 & 2, 2, 2, 2, 1, 1 & 15 & 16\\ \hline - & \mbox{Total} & 126 & -\\ \hline \end{array}$$

In every class, the denominator $n_1!\cdots n_6!$ is the same. Therefore

$$S=10!\times\left(\frac{6}{120}+\frac{30}{48}+\cdots+\frac{15}{16}\right)=16435440.$$

Then the probability is $$P=\frac{S}{6^{10}}=\frac{38045}{139968}\approx 0.2718.$$