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I'm not exactly sure how to do this using mathematical induction. Thanks for the help.

I'm trying to prove that the product rule is valid for some number of tasks $m$, such that $m$ is greater than two using induction and the product rule for two tasks as a given.

The product rule for a two-tasks procedure states that, there are $n_1n_2$ ways to do a procedure, when there are $n_1$ ways to do the first task and for each of these ways of doing the first task, there are $n_2$ ways to do the second task

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I can think of several possible variants, so you need to state exactly the product rule that you’re trying to prove. – Brian M. Scott Feb 22 '13 at 12:30
what is the product rule for $m$ tasks? – Emanuele Paolini Feb 22 '13 at 12:30
up vote 4 down vote accepted

Let $P(m)$ be the following statement:

If there are $n_k$ ways to perform task $T_k$ for $k=1,\dots,m$, then there are $n_1n_2\dots n_m$ ways to perform all $m$ tasks.

$P(1)$ is obvious, you’re given that $P(2)$ is true, and you’re to prove that $P(m)$ is true for all positive integers $m$. This is a natural setting for a proof by induction, and you’ve been handed the basis step for free. Thus, all you have to do is prove the induction step: if $P(m)$ is true, then so is $P(m+1)$. In words, if the product rule holds for $m$ tasks, then it holds for $m+1$ tasks. So assume that $P(m)$ is true for some $m\ge 2$; we’ll try to prove $P(m+1)$.

To that end suppose that we have $m+1$ tasks, and that task $T_k$ can be performed in $n_k$ ways, where $k=1,\dots,m+1$. We’d like to show that the entire set of $m+1$ tasks can be performed in $n_1n_2\dots n_mn_{m+1}$ ways.

HINT: Think of the combination of tasks $T_1$ through $T_m$ as a single very large task; call that task $T$. To complete tasks $T_1$ through $T_{m+1}$ you must complete the two tasks $T$ and $T_{m+1}$. You’re assuming that the product rule is true for $m$ tasks, so how many ways are there to perform task $T$? How many ways are there to perform the pair of tasks $T$ and $T_{m+1}$?

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Hint: Suppose you know that for a $(k-1)$-task procedure, there are $n_1 n_2 \cdots n_{k-1}$ ways to do the procedure. And that there are $n_k$ ways to do a $k$-th task. Then what does the product for two-task procedures tell you for the number of ways to do the two-task procedure: (1) perform the $(k-1)$-task procedure; (2) then perform the $k$-th task.

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