Take the tour ×  I'm lost on what this question is even asking:
If we make a sequence of $m$ choices for which there are $k_1$
possible ﬁrst choices, and for each way of making the ﬁrst $i −
1$ choices, there are $k_i$ ways to make the $i$th choice, then
in how many ways may we make our sequence of choices? (You need
not prove your answer correct at this time.)  share|improve this
question I think there are $k_2k_1$ ways to make the $2$nd
choice, and $k_3k_2k_1$ ways to make the third choice, ... and
$k_mk_{m-1}k_{m-2}\ldots k_2k_1$ ways to make the $m$th choice.
But I'm not confident enough of that to make it an answer. – 
Jeff Feb 17 '12 at 3:29 Jeff: No, there are $\Pi_{i=1}^nk_i$
ways to make the sequence of the first $n$ choices. For each
choice, there are exactly $k_i$ ways to make it. –  KReiser Feb
17 '12 at 3:32 add comment  1 Answer  up vote 4 down vote
accepted  $k_i$ is the number of ways to make the $i^{th}$
choice. So for example, if there are 6 ways to make the second
choice, then we have $k_2