9/7/2023 0 Comments Permutations and combinations![]() ![]() 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. Zwillinger, Daniel (Editor-in-Chief).ĬRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. 206, 2003.Example 1: How many 3 digit numbers can you make using the digits 1, 2 and 3 without repetitions? How many combinations are possible if customers are also allowed replacements when choosing toppings? For meats and cheeses this is now aĬombinations replacement or multichoose problem using the combinations with replacements equation:Ĭ R(n,r) = C(n+r-1, r) = (n+r-1)! / (r! (n - 1)!)įor meats, where the number of objects n = 5 and the number of choices r = 3, we can calculate eitherĬ R(5,3) = 35 or substitute terms and calculate combinations C(n+r-1, r) = C(5+3-1, 3) =Ĭalculating cheese choices in the same way, we now have the total number of possible options for each category atĪnd finally we multiply to find the total Now replacements are allowed, customers can choose any item more than once when they select their portions. In the previous calculation, replacements were not allowed customers had to choose 3 different meats and 2 different cheeses. 2 portions of one meat and 1 portion of another?. ![]() ![]() How many possible combinations are there if your customers are allowed to choose options like the following that still stay within the limits of the total number of portions allowed: Multiplying the possible combinations for each category we calculate:
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