# What is meant by $k_i$ in this combinatorics problem?

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.)

-
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

$k_i$ 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 $k_2=6$.

The question is asking how many ways we can make all of our choices if we can make some number of choices, $k_i$ in this case, the $i^{th}$ time we need to make a choice. Assuming none of these choices influence the others, then we have

$$k_1k_2\cdots k_n=\prod_{i=1}^{n} k_i$$ total ways to make all of our choices.

An example: if we need to make a sandwich, picking a sandwich bread from either rye, white, or wheat, picking a meat from beef, turkey, ham, or chicken, and picking a cheese from cheddar or pepperjack, we would have $k_1=3$, $k_2=4$, and $k_3=2$. This gives us a total of $k_1k_2k_3=3\cdot 4\cdot2=24$ ways to make a sandwich.

-