I NEED HELPP!!! PLEASE

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

Oksana_A [137]

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!

4 0

3 years ago

A sphere has a radius of 5.3 ft. What is the volume of the sphere to the nearest tenth? use 3.14 for pi. Question 3 options:

Alenkinab [10]

Your answer would be c
623.3 ft^3

5 0

3 years ago

Shelly is 3 years older than Michele. Four years ago the sum of their age was 67. Find the age of each person.

lawyer [7]

total of ages now is 67 + 4 + 4 = 75

S=shelly

M= Michelle

Michelle is 3 yrs younger than Shelly so use S-3

S +M = 75

replace M with S-3 to get:

S +S-3=75

2S=78

s=39

So shelly is 39

Michelle is 3 years younger = 39-3 = 36

4 0

3 years ago

Read 2 more answers

susan made 4 dozen cupcakes she placed them in boxes containing 4 cupcakes and sold each box for 8.75. how much money did she ea

Natasha_Volkova [10]

Answer: Susan has 105 dollars

Step-by-step explanation:

dozen is 12, 12 times 4 is 48, 48 divided by 4 is 12, 12 times 8.75 is 105

6 0

2 years ago

What is equivalent to 3 to the negative 2nd power

VladimirAG [237]

3 to the negative second power is written like this: $3^{-2}$ .

When simplifying an expression with a negative exponent, you take a positive version of the exponent, in this case that would be 2, and apply that to the base.
After doing that, we have 9.
The next step is to put one over that number.
In this case, now after doing that the answer is $\frac{1}{9}$ .

Hope this helps!

6 0

3 years ago