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Show transcribed image text Distinguishable permutations N! (n!(n2!)… (nk!) In class, Dr. Hom talked about distinguishable outcomes and how one needs to 'correct' probabilities that assume indistinguishability. The idea behind this relates to the concept of "permutations." For example, for 5 string of coin tosses (H = heads, T= tails), the probability that one would get "HHHHH" is (1/2)^5. Not swapping of Hs is going to change what it looks like, so these are indistinguishable. However, if one tosses 4 Hs and 11 than the "tossing sequence" can be either: "THHHH", "HTHHH", "HHTHH", "HHHTH", "HHHHT" so on needs to add up these probabilities (1/2)^5 for each), which is equivalent to multiplying by 5 in this case. (N = ny + N2 + … + nk) It turns out, mathematicians have figured out that the factor to multiply by is given by the general equation (known as the "distinguishable permutation formula") in the attached image, where N is the total number of sequence positions, "!" is the factorial operation (see: https://goo.gl/e4dxBG), and nl, n2,…, nk are the number of each type of different outcome (for a total of k different outcomes) and N=nl+n2+…+nk. In our second heads-tails example above, k=2 (Hor T), and nl = 4 (for 4 Hs), and n2 = (1 (for 1T). So the factor to multiply is: N! / (ni! n2!) = 5!/(1! 4!) = 5. Given this, what is the probability of rolling: two Is, three 3s, two 4s and two 6s with 9 fair dice? Enter your answer as a decimal, to 5 decimal digits. (Make sure you memorize and know how to use this "distinguishable permutation formula" with probabilities for exams!) Numeric Answer: Correct Answer 0.00075

Distinguishable permutations N! (n!(n2!)… (nk!) In class, Dr. Hom talked about distinguishable outcomes and how one needs to 'correct' probabilities that assume indistinguishability. The idea behind this relates to the concept of "permutations." For example, for 5 string of coin tosses (H = heads, T= tails), the probability that one would get "HHHHH" is (1/2)^5. Not swapping of Hs is going to change what it looks like, so these are indistinguishable. However, if one tosses 4 Hs and 11 than the "tossing sequence" can be either: "THHHH", "HTHHH", "HHTHH", "HHHTH", "HHHHT" so on needs to add up these probabilities (1/2)^5 for each), which is equivalent to multiplying by 5 in this case. (N = ny + N2 + … + nk) It turns out, mathematicians have figured out that the factor to multiply by is given by the general equation (known as the "distinguishable permutation formula") in the attached image, where N is the total number of sequence positions, "!" is the factorial operation (see: https://goo.gl/e4dxBG), and nl, n2,…, nk are the number of each type of different outcome (for a total of k different outcomes) and N=nl+n2+…+nk. In our second heads-tails example above, k=2 (Hor T), and nl = 4 (for 4 Hs), and n2 = (1 (for 1T). So the factor to multiply is: N! / (ni! n2!) = 5!/(1! 4!) = 5. Given this, what is the probability of rolling: two Is, three 3s, two 4s and two 6s with 9 fair dice? Enter your answer as a decimal, to 5 decimal digits. (Make sure you memorize and know how to use this "distinguishable permutation formula" with probabilities for exams!) Numeric Answer: Correct Answer 0.00075