(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, 3, 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 \(0 \le a_1 \le a_2 \le a_3 \le \dots \le a_k \le n\).

mathmathj28 Feb 25, 2020

#1**+1 **

https://web2.0calc.com/questions/a-count-the-number-of-quadruples-a-b-c-d-of-nonnegative

Guest Feb 25, 2020

#2**+1 **

\(\text{Any set of 4 distinct integers can be put in strictly ascending order}\\ \text{We can choose $\dbinom{13}{4}=715$ sets of distinct integers in $[0,12]$}\)

\(\text{b) is the same idea as (a) but now the integers don't have to be distinct}\\ \text{we have a few flavors of how the selected integers might appear}\\ \text{i) 4 distinct integers, ii) 1 pair, ii) 2 pairs, iv) 3 of a kind, v) 4 of a kind}\\ \text{For all these flavors there is only a single way they can be distinctly sorted in ascending order}\\ \text{so $N=\dbinom{13}{4}+13 \dbinom{12}{2} + \dbinom{13}{2}+13\cdot 12 + 13 = 1820$}\\ \text{where each of those terms corresponds to the number of each flavor that occurs}\)

you can do (c)

Rom Feb 25, 2020