+0  
 
0
54
1
avatar

(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 \(0 \le a_1 \le a_2 \le a_3 \le \dots \le a_k \le n\)

 Jan 12, 2020
 #1
avatar
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.

 Jan 12, 2020

20 Online Users