(a) Count the number of quadruples (a,b,c,d) of nonnegative integers such that 0 \le a < b < c < d \le 12.

(b) For this part, we want to count the number of quadruples (a,b,c,d) of nonnegative integers such that 0 \le a \le b \le c \le d \le 12. Here, some of a, b, c, and d can be equal to each other, so the answer will be different from part (a). Each value a, b, c, d must be between 0 and 12 inclusive. One idea is to count how many times each number appears. For example, suppose (a,b,c,d) = (1,3,8,8). Then we can make a table that counts how many times each number appears among a, b, c, and d:

Use this idea to find the number of quadruples (a,b,c,d). (If you come up with a different approach, you are free to use it.)

(c) In general, find the number of k-tuples (a_1, a_2, a_3, ...., a_k) of nonnegative integers such that

I need help with this question, I don't get it. Can someone please explain, I'm not trying to cheat, I also want to know how this problem is done.

Guest Feb 27, 2020

#1**0 **

(a) We can choose four numbers from 1, 2, 3, ..., 12. Then a is the lowest, b is the second-lowest, c is the third-lowest, and is the fourth-lowest, so the number of ways of choosing a,b,c,d is C(12,4) = 495.

(b) There are 12 ways of choosing a. If a = 12, then there are 12 ways of choosing b. If a = 11, then there are 11 ways of choosing b. This pattern continues, so the number of ways of choosing b is 12 + 11 + 10 + ... + 1.

If b = 11, then there are 11 ways of choosing c. If b = 10, then there are 10 ways of choosing c. This pattern continues, so the number of ways of choosing c is 11 + 10 + 9 + ... + 1.

If c = 10, then there are 10 ways of choosing d. If c = 9, then there are 9 ways of choosing d. This pattern continues, so the number of ways of choosing d is 10 + 9 + 8 + ... + 1.

So the number of ways of choosing a,b,c,d is 12(12 + 11 + ... + 1)(11 + 10 + 9 + .. + 1)(10 + 9 + 8 + ... + 1) = 12*78*66*55 = 3397680.

(c) There are n ways of choosing a_1. If a_1 = n, then there are n ways of choosing a_2. If a_1 = n - 1, then there are n - 1 ways of choosing a_2. This pattern continues, so the number of ways of choosing a_2 is n + (n - 1) + ... + 1 = n(n + 1)/2.

We can take the formula for part (b) and make it for n and k. The number of ways of choosing a_1,a_2,...,a_k is n * n(n + 1)/2 * (n - 1)n/2 * ... * (n - k)(n - k + 1)/2.

Guest Feb 27, 2020