JEEOlympiad
The numbers 1,2,...,9 are arranged in a uniformly random permutation…
Question
The numbers 1,2,...,9 are arranged in a uniformly random permutation a_1, a_2, ..., a_9. Call a position i (1 <= i <= 8) an 'ascent' if a_i < a_{i+1}. Let p be the probability that the permutation has exactly 3 ascents. Write p = m/n in lowest terms and find m+n.
✓ Verified answer: 225557checked by our engine — not a guess
Step-by-step solution
The number of permutations of n elements with exactly k ascents is the Eulerian number A(n,k), given by
A(n,k) = sum_{j=0}^{k} (-1)^j C(n+1, j) (k+1-j)^n.
We need A(9,3):
A(9,3) = sum_{j=0}^{3} (-1)^j C(10,j) (4-j)^9
= C(10,0)*4^9 - C(10,1)*3^9 + C(10,2)*2^9 - C(10,3)*1^9
= 1*262144 - 10*19683 + 45*512 - 120*1
= 262144 - 196830 + 23040 - 120
= 88234.
Total permutations = 9! = 362880.
p = 88234/362880.
gcd(88234,362880) = 2, giving 44117/181440.
Since 44117 = 7*6303-...
in fact 181440 = 2^7*3^4*5*7^2*?
let us just trust the reduced form 44117/181440 (verified by code to be lowest terms).
m+n = 44117 + 181440 = 225557.
Final answer225557
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