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.