(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:\(\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \end{array} \)

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 \((a_1, a_2, a_3, \dots, a_k)\) of nonnegative integers such that \(0 \le a_1 \le a_2 \le a_3 \le \dots \le a_k \le n.\)

Guest Jan 14, 2020

#1**+1 **

More than TEN of them ...Thanx guest...... I may have to look at this for a while longer !

ElectricPavlov Jan 14, 2020

edited by
ElectricPavlov
Jan 14, 2020

edited by ElectricPavlov Jan 14, 2020

edited by ElectricPavlov Jan 14, 2020

edited by ElectricPavlov Jan 14, 2020

edited by ElectricPavlov Jan 14, 2020

#4**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 Jan 14, 2020