2ARE METHODS TO SOLVE CERTAIN TYPES OF WORD PROBLEMS. PERMUTATIONSANDCOMBINATIONSARE METHODS TO SOLVE CERTAIN TYPES OF WORD PROBLEMS.
3PERMUTATIONS AND COMBINATIONS BOTHPERMUTATIONS AND COMBINATIONSUSE A COUNTING METHOD CALLED FACTORIAL.
4A FACTORIAL is a counting method that uses consecutive whole numbers as factors. The factorial symbol is !Examples 5! = 5x4x3x2x1= 1207! = 7x6x5x4x3x2x1= 5040
5First, we’ll do some permutation problems. Permutations are “arrangements”.
6Let’s do a permutation problem. 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 CBASix arrangements or 3! = 3x2x1 = 6
7In 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.
81. How many ways can 4 books be arranged on a shelf? You try this one:1. How many ways can 4 books be arranged on a shelf?4! or 4x3x2x1 or 24 arrangementsHere are the 24 different arrangements:ABCD ABDC ACBD ACDB ADBC ADCBBACD BADC BCAD BCDA BDAC BDCACABD CADB CBAD CBDA CDAB CDBADABC DACB DBAC DBCA DCAB DCBA
9Now we’re going to do 3 books on a shelf again, but this time we’re going to choose them from a group of 8 books.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.
10If we were looking for different arrangements for all 8 books, then we would do 8! But we only want the different arrangements for groups of 3 out of 8, so we’ll do a partial factorial,8x7x6=336
11Try these:1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there?10x9x8x7x6 or
122. Choose 4 books from a group of 7 and arrange them on a shelf 2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there?7x6x5x4 or
13Now, we’ll do some combination problems. Combinations are “selections”.
14There are some problems where the order of the items is NOT important. These are called combinations.You are just making selections, not making different arrangements.
15Example: 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.
16Because we’re not going to use all the possible 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.)Answer:10 committees5x4x33x2x1
17Now, you try.1. How many possible committees of 2 people can be selected from a group of 8?8x72x1or 28 possible committees
182. How many committees of 4 students could be formed from a group of 12 people? 12x11x10x94x3x2x1or 495 possible committees
19CIRCULAR PERMUTATIONS When items are in a circular format, to find the number of different arrangements, divide:n! / n
20Six students are sitting around a circular table in the cafeteria Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there?6! 6 = 120