9/1/2023 0 Comments Permutations and combinations![]() ![]() 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 different committee can be formed from the group? A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls.How many triangles can you make using 6 non collinear points on a plane?.In how many ways can you select a committee of 3 students out of 10 students?.In how many ways can you arrange 5 different books on a shelf?.How many 6 letter words can we make using the letters in the word LIBERTY without repetitions?.How many 3 digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions?.How many 4 digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions?.There is nothing that indicates that the order in which the team members are selected is imoportant and therefore it is a combination problem. 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.ģ C 2 = 3! / = 6 / = 3 (problem of points and lines solved above in example 6)ĥ C 5 = 5! / = 5! / = 5! / = 1 (there is only one way to select (without order) 5 items from 5 items and to select all of them once!)Įxample 8:We need to form a 5 a side team in a class of 12 students. This is a combination problem: combining 2 items out of 3 and is written as follows: 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. The lines are: AB, BC and AC 3 lines only. There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB. If we proceed as we did with permutations, we get the following pairs of points to draw lines. The problem is to select 2 points out of 3 to draw different lines. The number of words is given byĮxample 6: How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? Solution: 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. 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 formula is given byĦ P 5 = 6! / (6 - 5)! = 6×5×4×3×2×1 / 1! = 720Ĥ P 4 = 4! / (4 - 4)! = 4! / 0! = 4! = 4×3×2×1 = 24 (We now understand the need to define 0! = 1)Įxample 5: How many 3 letter words can we make with the letters in the word LOVE? 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. The most important idea in permutations is that order is important. ![]() The above problem is that of arranging 2 digits out of 4 in a specific order. Using the counting principle, the number of 2 digit numbers that we can make using 4 digits is given by This time we want to use 2 digits at the time to make 2 digit numbers.įor the first digit we have 4 choices and for the second digit we have 3 choices (4 - 1 used already). The number of permutations of n objects taken r at a time is determined by the following formula: P ( n, r) n ( n r) n is read n factorial and means all numbers from 1 to n multiplied e.g. Hence the number of words is given byĮxample 3: How many 2 digit numbers can you make using the digits 1, 2, 3 and 4 without repeating the digits? One could say that a permutation is an ordered combination. Solution: 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. In general n! is read n factorial and is given byĮxample 2: How many different words can we make using the letters A, B, E and L ? There is a special notation for the product 3 × 2 × 1 = 3! and it is read 3 factorial. The total number of 3-digit numbers is given by Using the counting principle, we can say: ![]() We have 3 choices for the first digit, 2 choices for the second digit and 1 choice for the third digit. We can make 6 numbers using 3 digits and without repetitions of the digits. Method (1) listing all possible numbers using a tree diagram. Now here are a couple examples where we have to figure out whether it is a permuation or a combination.Example 1: How many 3 digit numbers can you make using the digits 1, 2 and 3 without repetitions? If the order of the items is not important, use a combination. If the order of the items is important, use a permutation. Note: The difference between a combination and a permutation is whether order matters or not. There are 286 ways to choose the three pieces of candy to pack in her lunch. ![]()
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