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Kobotan [32]
3 years ago
14

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

e written in decreasing order but are not necessarily distinct? In other words, how many 5-tuples of integers (h, i, j, k, m) are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1?
Mathematics
1 answer:
Oksana_A [137]3 years ago
4 0

Answer:

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

 In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

{n \choose 5 } = \frac{n!}{5!(n-5)!}

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 {n \choose 4} .

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is 4 * {n \choose 4} .

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is {n \choose 3} .

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have {3 \choose 2 } = 3  possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of 6 * {n \choose 3}

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of {n \choose 2}  possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

4 * {n \choose 2}

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

I hope that works for you!

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Find the quartiles for these data values:
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<h3>Answer: Choice B</h3>

Q1 = 7, Q2 = 10, Q3 = 13.5

=============================================================

Explanation:

Start with {5,7,7,8,10,11,12,15,17}

Notice how this data set is already sorted for us from smallest to largest.

Cross off the first and last items to get {7,7,8,10,11,12,15}

Repeat the last step to get this smaller set {7,8,10,11,12}

Repeat again: {8,10,11}

Repeat one more time: {10}

The 10 is at the very center, so it is the median aka the value of Q2.

-------------

An alternative way to get the median is to follow these steps:

There are n = 9 numbers in the original set before we crossed off any items. The middle number is in slot 5 because n/2 = 9/2 = 4.5 rounds up to 5. The value in the fifth slot is 10, so 10 is the median.

There are 4 items below the median and 4 items above it, giving n = 4+1+4 = 9 items total.

-------------

Next, break the data set into two smaller groups

L = lower set = every value below the median

L = {5, 7, 7, 8}

U = upper set = every value above the median

U = {11, 12, 15, 17}

The median itself is in neither set L nor set U.

The median of set L is (7+7)/2 = 7, so this is the value of Q1

The median of set U is (12+15)/2 = 13.5 which is the value of Q3

-------------

Summary:

Q1 = 7

Q2 = 10 (aka the median of the original set)

Q3 = 13.5

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