Connecticut as a permutation, 1 example would be c1,o,n1,n2,e,c2,t1,I,c3,u,t2. 10 items can form 10! permutations. Considering c1=c2=c3, n1=n2, t1=t2, let’s look at the c all equal condition: 3cs existing one in front of or behind another may form c1, c2, c3; c1, c3, c2; c2,c1,c3… totally 6 different ôkkkiojobvarieties
It’s True.. you take the “extreme” variable from each proportion and cross multiply them before setting them equal to the product of the two “means” (which are just the other 2 numbers in the proportions).
