% 7 個(gè)人進(jìn)行了編號(hào)，1號(hào)和2號(hào)是新冠患者，3，4，5，6，7是健康的

%

a = 1:7;

V= perms(a);

[n,~] = size(V);

% 恰好第5次把兩個(gè)患者都檢測(cè)出來(lái)的情況進(jìn)行計(jì)數(shù)

s5 =0;

for i=1:n

???if (V(i,5) == 1 || V(i,5) == 2 ) && V(i,6) > 2 && V(i,7) >2

??????% 1或者2號(hào)排在第5位，第6位和第7位是健康者，這就是一個(gè)恰好第5次檢測(cè)出兩個(gè)患者的情況

??????s5 = s5+1;

???end

end

% 得出恰好第五次檢測(cè)出兩個(gè)患者的概率

s5/factoral(7)

% 把一個(gè)向量的所有排列都給出來(lái)

% 例 perms ([1, 2, 3])

%??3??2??1

%??3??1??2

%??2??3??1

%??2??1??3

%??1??3??2

%??1??2??3

%

function A = perms (v)

?if (nargin < 1)

???help perms

??????return

?end

?v = v(:).';

?if (isnumeric (v) || ischar (v))

???v = sort (v, "ascend");

?end

?n = numel (v);

?if (n < 4)???% 對(duì)于n比較小時(shí)，單獨(dú)處理

???switch (n)

?????case 0

???????A = reshape (v, 1, 0);

?????case 1

???????A = v;

?????case 2

???????A = [v([2 1]);v];

?????case 3

???????A = v([3 2 1; 3 1 2; 2 3 1; 2 1 3; 1 3 2; 1 2 3]);

???end

?else

???v = v(end:-1:1);

???n = n-1;

???idx = zeros (factorial (n), n);

???idx(1:6, n-2:n) = [1, 2, 3;1, 3, 2;2, 1, 3;2, 3, 1;3, 1, 2;3, 2, 1]+(n-3);

???f = 2;???% jump-start for efficiency with medium n

???for j = 3:n-1

?????b = 1:n;

?????f = f*j;

?????perm = idx(1:f, n-(j-1):n);

?????????tmp = n-j:n;

?????????tmpA = tmp(ones(f,1),:);

?????idx(1:(j+1)*f, n-j) = tmpA(:);

?????for i=0:j

???????b(i+n-j) = b(i+n-j) - 1;

???????idx((1:f)+i*f, n-(j-1):n) = b(perm);

?????end

???end

???n = n + 1;

???f = f*(n-1);

??????tmpC = v(1);

???A = tmpC(ones (factorial (n), n));

??????tmpD = v(ones(f,1),:);

???A(:,1) = tmpD(:);

???for i = 1:n

?????b = v([1:i-1 i+1:n]);

?????A((1:f)+(i-1)*f, 2:end) = b(idx);

???end

?end

end