1. The problem statement, all variables and given/known data RollingDice. Supposethat 12dice are to be rolled.We shall determine the probability p that each of the six different numbers will appear twice. Each outcome in the sample space S can be regarded as an ordered sequence of 12 numbers, where the ith number in the sequence is the outcome of the ith roll. Hence, there will be 6^12 possible outcomes in S, and each of these outcomes can be regarded as equally probable. The number of these outcomes that would contain each of the six numbers 1, 2,...,6 exactly twice will be equal to the number of differentpossiblearrangementsofthese12elements.Thisnumbercanbedetermined byevaluatingthemultinomialcoefﬁcientforwhichn=12,k=6,andn1=n2=...= n6=2. Hence, the number of such outcomes is 12 choose 2, 2, 2, 2, 2, 2= 12!/ (2!)6 , and the required probability p is p=12!/ (2!(6)(6^12)) =0.0034 2. Relevant equations Multinomial coefficient. 3. The attempt at a solution Here, the book uses multinomial coefficients to compute the number of outcomes for the given event. The thing is that I really think by heart that the solution is wrong. They are counting the number of outcomes of the sample space using permutation and the number of outcomes of the given event using combination. I really think that the right answer for the number of outcomes for the event is just 12! or another approach to solve this problem using multinomial coefficients is to change the counting method for the sample space, so it would be 72 choose 12. 72 would the total number of faces that will be distributed randomly in a set of 12 units.