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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