Answer
See step by step answer for solution.
Work Step by Step
To prove- P( m) , the sum rule for m tasks, which says that if tasks $T_{1}$, $T_2 , ... , T_m$ can be done in $n_1$ , $n_2$. . .. , $n_m$ ways, respectively, and no two of them can be done at the same time, then there are
$n_1$ + $n_2$+ . .. + $n_m$ ways to do one of the tasks.
Basis step- Take m = 2, and that has already been given.
Inductive step- Assume that P(m) is true, and we want to prove P(m + 1).
There are m + 1 tasks, no two of which can be done at the same time; we want to do one of them.
Either we can choose one from among the first m, or we choose the task $T_{m+1}$ · By the sum rule for two tasks, the number of ways we can do this is $n + n_{m+1}$ . where n is the number of ways we can do one of the tasks among the first m.
But by the inductive hypothesis n = $n_1$ + $n_2$+ . .. + $n_m$. Therefore the number of ways we can do one of the m + 1 tasks is
$n_1$ + $n_2$+ . .. + $n_m$+ $n_{m+1}$ , as desired.
