What does 9P2 mean, and how would one solve it?
We are studying "Sequences, Series, and Probability" and it likely related to binomial theorem and pascals triangle. I've a test tomorrow morning, and if I can't figure this out soon, I'm likely to fail.
$\endgroup$ 32 Answers
$\begingroup$$P(n,k)$ is the number of ways to create an ordered sample of size $k$ from a list of size $n$. It is easy to show that $$P(n,k) = {n!\over (n-k)!}.$$ Can you develop the reason why?
$\endgroup$ 2 $\begingroup$${^9P_2}$ is the number of permutations of the selections of 2 items from 9 distinct items. That is, the number of ordered selections of 2 items pulled from 9.
This is related to the count of the combinations of 2 items selected from 9, by multiplying that by the count of the permutations of 2 items.
$$\require{cancel}\begin{align}{^9\mathrm P_2} & = {^9\mathrm C_2}\times 2! \\[1ex] & = \frac{9!}{\cancel{2!}\;(9-2)!}\times \cancel{2!} \\[1ex] & = \frac{9!}{7!}\\[1ex] & = 9\times 8\\[1ex] & = 72\end{align}$$
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