In an investigation into 3 different treatments, 8 individualswere selected. In

Question

In an investigation into 3 different treatments, 8 individualswere selected. In a random manner, 3 were selected from the 8 fortreatment #1, 2 were selected at random from thre remaining 5 fortreatment #2, and the remaining were given treatment #3. In howmany ways can this be done? 8*7*6*5*4*3*2*1? In an investigation into 3 different treatments, 8 individualswere selected. In a random manner, 3 were selected from the 8 fortreatment #1, 2 were selected at random from thre remaining 5 fortreatment #2, and the remaining were given treatment #3. In howmany ways can this be done? 8*7*6*5*4*3*2*1?

Explanation / Answer

This is a combination problem. The selection of r itemsfrom a population of n is: C(n:r) =(n!)/{(r!)*[(n-r)!]} where n! = (n)*(n-1)*(n-2) ... (2)*(1) Example: 6! = 6*5*4*3*2*1 = 720 You will use the combination function two times: C(8:3) = (8!)/[(3!)*(5!)] = (8*7*6*5!)/[(6)*(5!)] = 56 C(5:2) = (5!)/[(2!)*(3!)] = (5*4*3!)/[(2)(3!)] = 10 there are (56)*(10) = 560 different ways to mix thepopulation using these particular combinations.

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