A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. For example, a phase I test of a specific drug involved only 5 subjects. Assume that we want to treat 5 healthy humans with this new drug and we have 10 suitable volunteers available. Complete parts (a) through (c) below. a. If the subjects are selected and treated in sequence, so that the trial is discontinued if anyone displays adverse effects, how many different sequential arrangements are possible if 5 people are selected from the 10 that are available? Choose the correct answer below. A. 30,240 C. 120 D. 252 b. If 5 subjects are selected from the 10 that are available, and the 5 selected subjects are all treated at the same time, how many different treatment groups are possible? There are \( \square \) different treatment groups possible.

To solve the problems, we will use combinatorial mathematics. ### Part (a) If the subjects are selected and treated in sequence, we need to consider the arrangements of 5 subjects selected from 10. This is a permutation problem because the order in which the subjects are treated matters. The number of ways to arrange \( r \) subjects from \( n \) subjects is given by the formula for permutations: \[ P(n, r) = \frac{n!}{(n - r)!} \] In this case, \( n = 10 \) and \( r = 5 \): \[ P(10, 5) = \frac{10!}{(10 - 5)!} = \frac{10!}{5!} = 10 \times 9 \times 8 \times 7 \times 6 = 30,240 \] So the answer for part (a) is: **A. 30,240** ### Part (b) If 5 subjects are selected from the 10 available and treated at the same time, we are looking for combinations, not permutations, because the order does not matter. The number of ways to choose \( r \) subjects from \( n \) subjects is given by the combination formula: \[ C(n, r) = \frac{n!}{r!(n - r)!} \] In this case, \( n = 10 \) and \( r = 5 \): \[ C(10, 5) = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \] So the answer for part (b) is: There are **252** different treatment groups possible.