The Number of Choices for Customer Queue Numbers

A bank security needs to print several customer queue numbers consisting of three digits. However, there is a restriction that the queue number cannot contain the same number formed from 0, 2, 2, and 3. In this article, we will explore the number of choices for queue numbers that can be made under this restriction. To solve this problem, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, we want to find the number of permutations of three digits without repetition, where the digits cannot be 0, 2, 2, or 3. First, let's consider the total number of possible permutations of three digits without any restrictions. Since there are 10 digits (0-9) to choose from, we have 10 choices for the first digit, 9 choices for the second digit (since we cannot repeat the first digit), and 8 choices for the third digit (since we cannot repeat the first or second digit). Therefore, the total number of permutations without any restrictions is 10 x 9 x 8 = 720. Now, let's consider the number of permutations that contain the same number formed from 0, 2, 2, and 3. Since there are two 2's in the restriction, we need to consider the number of permutations with two 2's and the number of permutations with three 2's separately. For the number of permutations with two 2's, we can fix one of the 2's in the first digit and then permute the remaining two digits. We have 3 choices for the first digit (0, 2, or 3), 2 choices for the second digit (0 or 3), and 1 choice for the third digit (the remaining digit). Therefore, the number of permutations with two 2's is 3 x 2 x 1 = 6. For the number of permutations with three 2's, we can fix one of the 2's in the first digit, one in the second digit, and permute the remaining digit. We have 3 choices for the first digit (0, 2, or 3), 2 choices for the second digit (0 or 3), and 1 choice for the third digit (the remaining digit). Therefore, the number of permutations with three 2's is 3 x 2 x 1 = 6. To find the number of choices for queue numbers that can be made under the given restriction, we subtract the number of permutations with two 2's and the number of permutations with three 2's from the total number of permutations without any restrictions. Therefore, the number of choices for queue numbers is 720 - 6 - 6 = 708. In conclusion, there are 708 choices of queue numbers that can be made by the bank security under the given restriction.