Permutations and Combinations Problems

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Maryann Mitchell
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1 Permutations and Combinations Problems Permutations and combinations are used to solve problems. Factorial Example 1: How many 3 digit numbers can you make using the digits 1, 2 and 3 without method (1) listing all possible numbers using a tree diagram. We can make 6 numbers using 3 digits and without repetitions of the digits. method (2) counting: LOOK AT THE TREE DIAGRAM ABOVE. We have 3 choices for the first digit, 2 choices for the second digit and 1 choice for the third digit. Using the counting principle, we can say:

2 The total number of 3-digit numbers is given by 3 * 2 * 1 = 6 There is a special notation for the product 3 * 2 * 1 = 3! and it is read 3 factorial. In general n! is read n factorial and is given by n! = n*(n - 1)*(n - 2)*...*2*1 We also define 0! = 1. Example 2: How many different words can we make using the letters A, B, E and L? We have 4 choices for the first letter, 3 choices for the second letter, 2 choices for the third letter and 1 choice for the fourth letter. Hence the number of words is given by 4 * 3 * 2 * 1 = 4! = 24 Permutations Example 3: How many 2 digit numbers can you make using the digits 1, 2, 3 and 4 without repeating the digits? This time we want to use 2 digits at the time to make 2 digit numbers. For the first digit we have 4 choices and for the second digit we have 3 choices (4-1 used already). Using the counting principle, the number of 2 digit numbers that we can make using 4 digits is given by 4 * 3 = 12 The above problem is that of arranging 2 digits out of 4 in a specific order. This is also called permutating. The most important idea in permuatations is that order is important. When you use the digits 3 and 4 to make a number, the number 34 and 43 are different hence the order of the digits 3 and 4 is important. In general permutating r (2 digit in the above example) items out of a set of n (4 digits in the above example) items is written as n P r and the formuala is given by n P r = n! / (n - r)!

3 Example 4: Calculate 4 P 2 6 P 5 4 P 4 4 P 2 = 4! / (4-2)! = 24/2 = 12 6 P 5 = 6! / (6-5)! = 6*5*4*3*2*1 / 1! = P 4 = 4! / (4-4)! = 4! / 0! = 4! = 4*3*2*1 = 24 (We now understand the need to define 0! = 1) Example 5: How many 3 letter words can we make with the letters in the word LOVE? There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. Hence it is a permutation problem. The number of words is given by 4 P 3 = 4! / (4-3)! = 24 Combinations Example 6: How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permuatations, we get the following pairs of points to draw lines. AB, AC BA, BC CA, CB There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB. The lines are: AB, BC and AC ; 3 lines only. So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

4 This is a combination problem: combining 2 items out of 3 and is written as follows: n C r = n! / [ (n - r)! r! ] The number of combinations is equal to the number of permuations divided by r! to eliminates those counted more than once because the order is not important. Example 7: Calculate 3 C 2 5 C 5 3 C 2 = 3! / [ (3-2)!2! ] = 6 / [1 * 2] = 3 (problem of points and lines solved above in example 6) 5 C 5 = 5! / [(5-5)!5! ] = 5! / [0!5!] = 5! / [1 * 5!] = 1 (there is only one way to select (without order) 5 items from 5 items and to select all of them once!) Example 8:We need to form a 5 a side team in a class of 12 students. How many different teams can be formed? There is nothing that indicates that the order in which the team members are selected is imoportant and therefore it is a combination problem. Hence the number of teams is given by 12 C 5 = 12! / [ (12-5)!5! ] = 792 Problems with solutions 1. How many 4 digit numbers can we make using the digits 3, 6, 7 and 8 without 2. How many 3 digit numbers can we make using the digits 2, 3, 4, 5, and 6 without 1. How many 6 letter words can we make using the letters in the word LIBERTY without 2. In how many ways can you arrange 5 different books on a shelf? 3. In how many ways can you select a committee of 3 students out of 10 students? 4. How many triangles can you make using 6 non collinear points on a plane?

5 5. A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. How many different committee can be formed from the group? 6. In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. How many number plates can be formed if neither the digits nor the letters are repeated? Solutions 1. 4! = P 3 = P 6 = ! = C 3 = C 3 = C 3 * 12 c 4 = 59, P 4 * 26 P 3 = 47,174,400

Tree Diagrams and the Fundamental Counting Principle

Tree Diagrams and the Fundamental Counting Principle Objective: In this lesson, you will use permutations and combinations to compute probabilities of compound events and to solve problems. Read this knowledge article and answer the following: Tree Diagrams