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%I #35 Dec 11 2021 04:58:44
%S 1,1092,793884,480299820,261163522620,132358677731280,
%T 63798093049771080,29612552769907347240,13345042642324219106280,
%U 5872442544965392834838400,2533775368098060137659608000,1075256447734638237381213700800
%N Numerator Q of probability P = Q(n)/365^(n-1) that exactly two out of n people share the same birthday.
%C A 365-day year and a uniform distribution of birthdays throughout the year are assumed.
%H Patrice Le Conte, <a href="/A225852/a225852.pdf">Coincident Birthdays</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BirthdayProblem.html">Birthday Problem</a>.
%F P(n) = n!*Sum_{i=1..floor(n/2)} binomial(365, i)*binomial(365-i, n-2*i)/2^i.
%e a(3)=1092 because the probability that in a group of 3 people exactly two of them share the same birthday is (1/365^3)*3!*binomial(365,1)*binomial(364,1)/2 = (1/365^2)*3*364 = (1/365^2)*1092.
%t P[n_] := (n! Sum[ Binomial[365, i]*Binomial[365 - i, n - 2i] /2^i, {i, 1, Floor[n/2]}]/365); Table[ P[n], {n, 2, 13}] (* _Robert G. Wilson v_, Feb 09 2004 *)
%Y Cf. A014088, A091674 gives probabilities for two or more coincidences, A091715 gives probabilities for three or more coincidences.
%K frac,nonn
%O 2,2
%A _Hugo Pfoertner_, Feb 03 2004
%E More terms from _Robert G. Wilson v_, Feb 09 2004
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