Permutations and Combinations

2.5 STEP THREE: The Labeling Principle

Lesson materials located below the video overview.

In how many ways can we arrange the letters of the Swedish pop group name  ABBA?


Answer: \(\dfrac{4!}{2!2!} = \dfrac{24}{4} = 6\).


In how many ways can we arrange the letters of AABBBBA?


Answer: \(\dfrac{7!}{3!4!}\).


In how many ways can we arrange the letters of AAABBBBCCCCCC?


Answer: \(\dfrac{13!}{3!4!6!}\).



Let’s look at this third problem and phrase it in a different way:


Mean Mr. Muckins has a class of 13 students. He has decided to randomly assign the grade of A to three students, the grade of B to four students, and the grade of C to six students. In how many ways could he assign these labels?


Answer: Let’s imagine all thirteen students are in a line.

Here’s one way he can assign labels:


Here’s another way:


and so on.


We see that this labeling problem is just the same problem as rearranging letters. The answer must be \(\dfrac{13!}{3!4!6!}\).


Of 10 people in an office 4 are needed for a committee. How many ways?


Answer: Imagine the 10 people standing in a line. We need to give out labels. Four people will be called “ON” and six people will be called “LUCKY.” Here is one way to assign those labels:


We see that this is just a word arrangement problem. The answer is:

\(\dfrac{10!}{4!6!}=\dfrac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1}  = 10 \cdot 3 \cdot 7 = 210\).


In general, we have …


THE LABELING PRINCIPLEEach of \(N\) distinct objects is to be given a label. If \(a\) of them are to have label “1,” \(b\) of them to have label “2,” and so on, then the total number of ways to assign labels is: 

\(\dfrac{N!}{a!b! \cdots z!}\).


That’s it. We are just arranging letters, with the letters being the names of the labels.


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