Rule of division (combinatorics)

In combinatorics, the rule of division is a counting principle. It states that there are $n/d$ ways to do a task if it can be done using a procedure that can be carried out in $n$ ways, and for each way $w$, exactly $d$ of the $n$ ways correspond to the way $w$. In a nutshell, the division rule is a common way to ignore "unimportant" differences when counting things.

Applied to Sets
In the terms of a set: "If the finite set $A$ is the union of n pairwise disjoint subsets each with $d$ elements, then $n = |A|/d$."

As a function
The rule of division formulated in terms of functions: "If $f$ is a function from $A$ to $B$ where $A$ and $B$ are finite sets, and that for every value $y ∈ B$ there are exactly $d$ values $x ∈ A$ such that $f (x) = y$ (in which case, we say that $f$ is $d$-to-one), then $|B| = |A|/d$."

Examples


Example 1

- How many different ways are there to seat four people around a circular table, where two seatings are considered the same when each person has the same left neighbor and the same right neighbor?


 * To solve this exercise we must first pick a random seat, and assign it to person 1, the rest of seats will be labeled in numerical order, in clockwise rotation around the table. There are 4 seats to choose from when we pick the first seat, 3 for the second, 2 for the third and just 1 option left for the last one. Thus there are 4! = 24 possible ways to seat them. However, since we only consider a different arrangement when they don't have the same neighbours left and right, only 1 out of every 4 seat choices matter.
 * Because there are 4 ways to choose for seat 1, by the division rule ($n/d$) there are $24/4 = 6$ different seating arrangements for 4 people around the table.

Example 2

- We have 6 coloured bricks in total, 4 of them are red and 2 are white, in how many ways can we arrange them?


 * If all bricks had different colours, the total of ways to arrange them would be $6! = 720$, but since they don't have different colours, we would calculate it as following:
 * 4 red bricks have $4! = 24$ arrangements
 * 2 white bricks have $2! = 2$ arrangements
 * Total arrangements of 4 red and 2 white bricks = $6!⁄4!2! = 15$.