Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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
2  
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

1 Answer 1

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 $k_2=6$.

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.