Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why do we divide by $k$ when counting $C(n,r)$?

For example, I have 4 different balls (A,B,C,D).

I want to pick 3 of them and not put back.

I know the formula and I got the answer is 4.

$$c_r^n = \frac{p_r^n}{r!}=4$$

  1. ABC

  2. ABD

  3. ACD

  4. BCD

But what's the idea of formula to divide 3 and also divide 2?

Thank you~

share|cite|improve this question
up vote 1 down vote accepted

Once you select the three balls, counting order (which is what you do with $P^n_r$), how many times will a particular choice of $3$ balls occur? There are $P^3_3$ ways of listing those three balls, so by taking order into account you are "counting" each selection of $3$ objects $P^3_3$ times. So you must divide by $P^3_3 = 3 = 3\times 2$ to get the right answer.

(Alternatively, you have $3$ ways of picking a "first" ball, $2$ ways of picking a "second" ball from the remaining ones, and $1$ way of picking the third ball from the remaining ones; this gives $3\times 2\times 1=3!$ repetitions).

The general formula is $$C^{n}{r} = \frac{P^n_r}{P^r_r}$$ because when counting with $P^n_r$, each selection of $r$ objects will be counted $P^r_r=r!$ times.

share|cite|improve this answer

Your Answer


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.