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I am trying to solve this question. The direct computation using the principle of inclusion/exclusion makes sense. However, my first attempt at solving this problem was using the complement calculation. For this problem, the complement case would be 0 of 6 values never appear, which in other words says that all 6 values appear. Based on this idea,

$$P(E) = 1 - \frac{\binom{n}{6} \ 6! \ 6^{n-6}}{6^{n}}.$$

Clearly, the answer using this approach doesn't match the direct approach using which we can obtain,

$$P(E) = 6 \left(\frac{5}{6}\right )^{n} - \binom{6}{2} \left(\frac{4}{6}\right )^{n} + \binom{6}{3} \left(\frac{3}{6}\right )^{n} - \binom{6}{4} \left(\frac{2}{6}\right )^{n} + \binom{6}{5} \left(\frac{1}{6}\right )^{n}.$$

Can someone please point out as to where I am going wrong? Thanks.

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    $\begingroup$ Your method over counts the cases in which all $6$ appear. Specifically, if, say, the $5$ shows up $7$ times, you count that pattern at least $7$ times. Note, for instance, that your expression is very negative if $n$ is large. $\endgroup$
    – lulu
    Commented Feb 7, 2020 at 22:49
  • $\begingroup$ For intuition, work this with a coin instead of a die. For that case the direct method is simple: there are only two possible patterns which are missing one of $H, T$. But try to apply your method . $\endgroup$
    – lulu
    Commented Feb 7, 2020 at 22:57
  • $\begingroup$ @lulu I have been thinking about your comments. Although the statement, "Specifically, if, say, the 5 shows up 7 times, you count that pattern at least 7 times." still doesn't make sense to me, I can see that the expression is negative for large n. Also, using a coin instead of a die makes it much easier to visualize this (I tried writing down the possibilities with n=3 and could see the over-counting). Do you think there is a way to count the number of possibilities using the direct method? $\endgroup$ Commented Feb 9, 2020 at 16:48
  • $\begingroup$ I do not see any method easier than Inclusion-Exclusion for the die problem. You can't easily use something like Stars and Bars, for instance, as some patterns are more likely than others. $\endgroup$
    – lulu
    Commented Feb 9, 2020 at 16:57
  • $\begingroup$ For my comment about $5$ showing up $7$ times: your method tries to single out a "special" occurrence of each of the digits. That's what the term $\binom n6$ does. However, given a pattern there is no way to tell which occurrence was special. Thus your method counts the pattern repeated, at least once for every occurence of a digit. $\endgroup$
    – lulu
    Commented Feb 9, 2020 at 16:59

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