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A permutation of the numbers 1,2,...,8 is chosen uniformly at random.…
Question
A permutation of the numbers 1,2,...,8 is chosen uniformly at random. Let p be the probability that the permutation has exactly 3 fixed points (positions i where the value equals i). Write p = m/n in lowest terms. Find m+n.
✓ Verified answer: 191checked by our engine — not a guess
Step-by-step solution
To have exactly 3 fixed points: first choose which 3 of the 8 positions are fixed in C(8,3) ways.
The remaining 5 elements must form a permutation with NO fixed points, i.e.
a derangement of 5 elements.
Number of derangements D(5) = 5!(1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!) = 120 - 120 + 60 - 20 + 5 - 1 = 44.
Favourable count = C(8,3) * D(5) = 56 * 44 = 2464.
Total permutations = 8! = 40320.
p = 2464/40320 = 11/180.
m+n = 11+180 = 191.
Final answer191
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