For this problem, we'll calculate the probabilty of getting all 12 numbers after rolling a 12sided die 35 times. (Incidentally, the 12 sided die shape is called a dodecahedron. You have probably seen this shape in the form of a plastic desk calendar with one month printed on each of the 12 faces.)
which equals 5.91 × 10^{37} This number represents all the possible results from rolling a 12 sided die In steps 2, 3, 4 and 5, we will determine how many of those 12^{35} rolls, will contain all 12 numbers.
STEP 2 11^{35} = 2.81 × 10^{36} 10^{35} = 1.00 × 10^{35} 9^{35} = 2.50 × 10^{33} 8^{35} = 4.06 × 10^{31} 7^{35} = 3.79 × 10^{29} 6^{35} = 1.72 × 10^{27} 5^{35} = 2.91 × 10^{24} 4^{35} = 1.18 × 10^{21} 3^{35} = 5.00 × 10^{16} 2^{35} = 3.44 × 10^{10} 1^{35} = 1
STEP 3 11 C 12 = 12 10 C 12 = 66 9 C 12 = 220 8 C 12 = 495 7 C 12 = 792 6 C 12 = 934 5 C 12 = 792 4 C 12 = 495 3 C 12 = 220 2 C 12 = 66 1 C 12 = 12 Basically, this is saying that 11 objects can be chosen from a set of 12 in 12 ways 10 objects can be chosen from a set of 12 in 66 ways 9 objects can be chosen from a set of 12 in 220 ways 8 objects can be chosen from a set of 12 in 495 ways 7 objects can be chosen from a set of 12 in 792 ways 6 objects can be chosen from a set of 12 in 934 ways 5 objects can be chosen from a set of 12 in 792 ways 4 objects can be chosen from a set of 12 in 495 ways 3 objects can be chosen from a set of 12 in 220 ways 2 objects can be chosen from a set of 12 in 66 ways 1 object can be chosen from a set of 12 in 12 ways
STEP 4 2.81 × 10^{36} × 12 = 3.37 × 10^{37} 1.00 × 10^{35} × 66 = 6.60 × 10^{36} 2.50 × 10^{33} × 220 = 5.51 × 10^{35} 4.06 × 10^{31} × 495 = 2.01 × 10^{34} 3.79 × 10^{29} × 792 = 3.00 × 10^{32} 1.72 × 10^{27} × 934 = 1.59 × 10^{30} 2.91 × 10^{24} × 792 = 2.31 × 10^{27} 1.18 × 10^{21} × 495 = 5.84 × 10^{23} 5.00 × 10^{16} × 220 = 1.10 × 10^{19} 3.44 × 10^{10} × 66 = 2.27 × 10^{12} 1 × 12 = 12
STEP 5
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