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# last time i posted it said it was invisible need help due today

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(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 14, 2020

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

Jan 14, 2020
edited by ElectricPavlov  Jan 14, 2020
edited by ElectricPavlov  Jan 14, 2020
edited by ElectricPavlov  Jan 14, 2020
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but what about 12,7,4,3? that counts

Guest Jan 14, 2020
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i figured it out. I will try to post the answer after my problem is graded or they will think im cheating

Guest Jan 14, 2020
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(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 14, 2020