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DedPeter [7]
3 years ago
12

Suppose ten students in a class are to be grouped into teams. (a) If each team has two students, how many ways are there to form

teams? (The ordering of students within teams does not matter, and the ordering of the teams does not matter.) (b) If each team has either two or three students, how many ways are there to form teams?
Mathematics
1 answer:
ValentinkaMS [17]3 years ago
6 0

Answer:

(a) There are 113,400 ways

(b) There are 138,600 ways

Step-by-step explanation:

The number of ways to from k groups of n1, n2, ... and nk elements from a group of n elements is calculated using the following equation:

\frac{n!}{n1!*n2!*...*nk!}

Where n is equal to:

n=n1+n2+...+nk

If each team has two students, we can form 5 groups with 2 students each one. Then, k is equal to 5, n is equal to 10 and n1, n2, n3, n4 and n5 are equal to 2. So the number of ways to form teams are:

\frac{10!}{2!*2!*2!*2!*2!}=113,400

For part b, we can form 5 groups with 2 students or 2 groups with 2 students and 2 groups with 3 students. We already know that for the first case there are 113,400 ways to form group, so we need to calculate the number of ways for the second case as:

Replacing k by 4, n by 10, n1 and n2 by 2 and n3 and n4 by 3, we get:

\frac{10!}{2!*2!*3!*3!}=25,200

So, If each team has either two or three students, The number of ways  form teams are:

113,400 + 25,200 = 138,600

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Meredith is playing games at an arcade to earn tickets that she can exchange for a prize. She has 250 tickets from a previous vi
Nataly_w [17]

Considering the definition of equation, the number of games Meredith need to play is 12.

<h3>What is an equation</h3>

An equation in mathematics is defined as an equality established between two expressions, in which there may be one or more unknown values, called unknowns, in addition to certain known data.

The unknown or unknowns represent the number (or numbers), if any, that make the equality true. This unknown number or numbers represent the solution of the equation. So when changing the unknown or unknowns by the solution, the equality must be true.

<h3>Solution in this case</h3>

Meredith is playing games at an arcade to earn tickets that she can exchange for a prize. This is represented by:

y=6x + 250

where:

  • x represents the number of games Meredith plays.
  • y represents the number of tickets she needs for the prize.

Meredith needs 322 tickets to exchange for the prize she wants. This is, y=322.

Replacing in the expression y=6x + 250:

322= 6x + 250

Solving taking into account that:

  • When a value you are adding is passed to another side of the equation, this value will be subtracting.
  • If a value is multiplying it goes to the other side of the equation by dividing.

322 - 250=6x

72= 6x

72÷6= x

<u><em>12= x</em></u>

Finally, the number of games Meredith need to play is 12.

Learn more with this problem:

brainly.com/question/15564756

#SPJ1

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2 years ago
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The original price was 90$ so you saved 11.25$
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Step-by-step explanation:

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The function f(x)=70n -400 models the profit of the instructor of a guitar class per month, where n is the number of students en
suter [353]
Given:
f(x) = 70n - 400   : equation for monthly profit.
n = number of students enrolled in the guitar class.

70 refers to the amount charged per student.
400 is the fixed expense made by the instructor.

For the instructor to have a profit, the product of 70n must be more than 400.

400/70 = 5.7 or 6

There must be at least 6 students enrolled in class for the instructor to generate profit.

f(x) = 70(6) - 400 = 420 - 400 = 20

1000 = 70n - 400
1000 + 400 = 70n
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1400/70 = n
20 = n

There must be 20 students enrolled in class for the instructor to earn 1000 in profit.

1000 = 70(20) - 400
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3 0
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