Why doesn't the multiplication rule work here?

I have the following question:

How many ways are there to select $$3$$ candidates from $$8$$ equally qualified recent graduates for openings in an accounting firm?

I was given the following solution to it: $$\binom{8}{3}$$, however, I got the answer: $$8 \times 7 \times 6$$ by doing the multiplication rule, I was wondering why my answer doesn't work. My initial reason is that in the multiplication rule there is the underlying assumption that order matters and here obviously order doesn't matter. However, I am posting this here as that is the first time I come to such a realization and I am not sure that my reasoning is correct, especially since that I never learned multiplication rule with the idea of order in my mind. Any clarification would be appreciated.

• Because picking Jack, Jill, and Kim is the same as picking Jill, Kim, and Jack. Mar 6 '21 at 11:43
• How many ways are there of picking exactly 2 items from $\{a,b,c,d\}$? Is it $(4 \times 3)$ or is it $\frac{4 \times 3}{2}$? Mar 6 '21 at 11:45
• But would it safe to ALWAYS assume that in the multiplication rules order doesn't matter? Mar 6 '21 at 11:50
• No! The safest course is to either take a College course in Combinatorics, or get a corresponding textbook for self-study. While intuition is crucial for Combinatorics, the intuition must be nurtured/grown/stretched by studying the material, either with a coach (i.e. teacher) or on your own. Mar 6 '21 at 12:01
• A concern I have with this wording is that asking for how many ways seems to suggest that order is included. Better would be something like: "How many selections of 3 candidates are possible from ...", where it is clearer that what is being counted are the possible selection results and not the way in which the selections are made. Mar 6 '21 at 12:01

I think your question seems to be when you can assume order doesn't matter, and when it actually does.

I'll give you two examples which might make it easier.

The first is like your own question. How many ways of choosing 3 people out of 8 people? You're just choosing people, and not assigning any special role or task to them to distinguish between individual members of your chosen triple. Here, use $$^8C_3$$.

The second is a variation of your question. How many ways of choosing 3 club office holders (namely President, Secretary and Treasurer) out of 8 people? From the wording, you're not just choosing three people, you're also assigning each to a particular role. In this case, order definitely matters since $$(a, b, c)$$ in the respective roles of President, Secretary and Treasurer is a very different selection from $$(c, b, a)$$ in those respective roles. So here you use $$^8P_3$$ (which is equal to $$8\cdot 7 \cdot 6$$.

Yes you are right. The formula for the 'choose' function is $$\frac{n!}{k!(n-k)!}$$ where k is the number of things you choose and n is the number of things you choose from. If you do $$8 \times 7 \times 6$$ you are working out the number of permutations (where order matters). But you're trying to work out the number of combination and therefore you need to exclude all combinations where the order is the same. E.g. abc is the same as acb = bac = bca = cab = cba

(n-k)! takes care of how many people to choose each time (aka 3). But k! allows you to neglect the order and make sure each combination is distinct from another.

If you did $$\frac{8!}{8-3!}$$ you'd get $$8 \times 7 \times 6$$ but that would only give you the number of permutations

You want combinations so you should do $$\frac{8!}{3!8-3!}$$

Hope this helps!

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. In particular, you can write $8 \times 7 \times 6$ to obtain $8 \times 7 \times 6$ or $8 \cdot 7 \cdot 6$ to obtain $8 \cdot 7 \cdot 6$. Mar 6 '21 at 11:56

$$\binom{8}{3}$$ is the same as $$\frac{8!}{3!5!}$$ which is $$\frac{8.7.6}{3!}$$, so you just need to divide by the extra $$3!$$

That's because you have overlapping choices by e.g choosing ABC, is the same as ACB,BAC,BCA,CAB and CBA, 6 ways to get the same ABC.

For example, let's say that the candiates' names are A, B, C, and so on. Choosing A,B,C is the same as choosing B,A,C or C,A,B and so on. There are $$3! = 6$$ ways to pick the same candiates, so the answer will be $$\frac{8\times7\times6}{3\times2\times1}$$ = $$56$$.
This is due to picking candidates to do for the same thing. You need to divide by the factorial of the number of elected only when you're picking the same things or people. If not, the multiplication rule works without needing to divide by something.