PERMUTATIONS AND COMBINATIONS. PERMUTATIONS AND COMBINATIONS ARE METHODS TO SOLVE CERTAIN TYPES OF WORD PROBLEMS. BOTH PERMUTATIONS AND COMBINATIONS USE A COUNTING METHOD CALLED FACTORIAL . A FACTORIAL is a counting method that uses consecutive whole numbers as factors.
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ARE METHODS TO SOLVE CERTAIN TYPES OF WORD PROBLEMS.
PERMUTATIONS AND COMBINATIONS
USE A COUNTINGMETHOD CALLEDFACTORIAL.
A FACTORIAL is a counting method that uses consecutive whole numbers as factors.
The factorial symbol is !
Examples 5!= 5x4x3x2x1
7! = 7x6x5x4x3x2x1
First, we’ll do some permutation problems.
Permutations are “arrangements”.
Let’s do a permutationproblem.
How many different arrangements are there for 3 books on a shelf?
Books A,B, and C can be arranged in these ways:
ABC ACB BAC BCA CAB CBA
Six arrangements or 3! = 3x2x1 = 6
In a permutation, the order of the books is important.
Each different permutation is a different arrangement.
The arrangement ABC is different from the arrangement CBA, even though they are the same 3 books.
1. How many ways can 4 books be arranged on a shelf?
4! or 4x3x2x1 or 24 arrangements
Here are the 24 different arrangements:
ABCD ABDC ACBD ACDB ADBC ADCB
BACD BADC BCAD BCDA BDAC BDCA
CABD CADB CBAD CBDA CDAB CDBA
DABC DACB DBAC DBCA DCAB DCBA
We’re going to have a lot more possibilities this time, because there are many groups of 3 books to be chosen from the total 8, and there are several different arrangements for each group of 3.
But we only want the different arrangements for groups of 3 out of 8, so we’ll do a partial factorial,
1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there?
10x9x8x7x6 or 30240
2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there?
7x6x5x4 or 840
Now, we’ll do some shelf. How many different combination problems.
Combinations are “selections”.
There are some problems where the shelf. How many different orderof the items is NOT important.
These are called combinations.
You are just making selections, not making different arrangements.
Example: shelf. How many different A committee of 3 students must be selected from a group of 5 people. How many possible different committees could be formed?
Let’s call the 5 people A,B,C,D,and E.
Suppose the selected committee consists of students E, C, and A. If you re-arrange the names to C, A, and E, it’s still the same group of people. This is why the order is not important.
Because we’re not going to use all the possible shelf. How many different combinations of ECA, like EAC, CAE, CEA, ACE, and AEC, there will be a lot fewer committees.
Therefore instead of using only 5x4x3, to get the fewer committees, we must divide.
(Always divide by the factorial of the number of digits on top of the fraction.)
Now, you try. shelf. How many different
1. How many possible committees of 2 people can be selected from a group of 8?
or 28 possible committees
2. How many committees of 4 students could be shelf. How many different formed from a group of 12 people?
or 495 possible committees
CIRCULAR PERMUTATIONS shelf. How many different
When items are in a circular format, to find the number of different arrangements, divide:
n! / n
Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there?
6! 6 = 120
Order Order cafeteria. How many different seating arrangements are there? is isimportant not important
Don’t Dodivide divide
THE END cafeteria. How many different seating arrangements are there?