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The problem is:

An experimenter is studying the effects of temperature, pres-sure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration.

a.If any particular experimental run involves the use of a single temperature, pressure, and catalyst, how many experimental runs are possible?

b.How many experimental runs are there that involve use of the lowest temperature and two lowest pressures?

c. Suppose that five different experimental runs are to be made on the first day of experimentation. If the five are randomly selected from among all the possibilities, so that any group of five has the same probability of selection, what is the probability that a different catalyst is used on each run?

I am stuck on part c. I know that the total number of experiments is, $3 \cdot 4 \cdot 5=60$; and so, the total number of ways of choosing five of the sixty experiments would be ${{60}\choose{5}}$. What I am having trouble with is finding the total number of sequence of five experiments where a catalyst is not repeated in each sequence. I thought, for the first experiment, there would be $5 \cdot 4 \cdot 3$, and for the second experiment, $4 \cdot 4 \cdot 3$, and for the third, $3 \cdot 4 \cdot 3$, and so on. However, this isn't correct. What am I doing wrong?

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Hmm, people just seemed to suddenly stop viewing this thread. Perhaps word got out that this is too boring of a problem. – Mack Feb 3 '13 at 18:32
up vote 0 down vote accepted

The problem is that you disregarded the order of the runs in the first calculation but not in the second. Either is OK, but you need to do it the same way in both calculations. The order accounts for a factor of $5!=120$, so your result for the probability should be $120$ times the correct value.

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So are you saying that I should have done a permutation, instead of a combination? Implying that the order in which you choose the experiments matters? – Mack Feb 3 '13 at 19:33
@Eli: To the contrary, I explicitly said that it's OK to do either. The error was not in taking order into account or not; it was in doing one in the one calculation and the other in the other calculation. If you do them consistently, it doesn't matter, since the factors of $5!$ cancel. – joriki Feb 3 '13 at 19:57
Well, I did $\LARGE \frac{5! \cdot 12^5}{P_{60, 5}}$, and this produces the correct answer. – Mack Feb 3 '13 at 20:16

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