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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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Parallel / Perpendicular Practice
deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

  1. Neither
  2. ║
  3. Neither
  4. ⊥
  5. ║
  6. Neither
  7. Neither
  8. Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; m_1 = -\dfrac{1}{m_2}

1. The given equations are in the slope and intercept form

\ y = 3 \cdot x + 1

The slope, m₁ = 3

y = \dfrac{1}{3} \cdot x + 1

The slope, m₂ = \dfrac{1}{3}

Therefore, the equations are <u>neither</u> parallel or perpendicular

  • Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

y = 5 \cdot x -\dfrac{7}{2}

Therefore, the two equations are <u>parallel</u>

  • ║

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

y = 2 -\dfrac{1}{2}  \cdot x

Slope, m₁ = -\dfrac{1}{2}

y = \dfrac{1}{2}  \cdot x - 2

Slope, m₂ = \dfrac{1}{2}

The slopes

Therefore, m₁ ≠ m₂

m_1 \neq -\dfrac{1}{m_2}

The lines are <u>Neither</u> parallel nor perpendicular

  • <u>Neither</u>

4. The given equations are;

2·y - x = 2

y = \dfrac{1}{2} \cdot   x +1

m₁ = \dfrac{1}{2}

y = -2·x + 4

m₂ = -2

Therefore;

m_1 \neq -\dfrac{1}{m_2}

Therefore, the lines are <u>perpendicular</u>

  • ⊥

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, y = \dfrac{3}{4} \cdot x + 3

Second equation, y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}

Therefore, m₁ = m₂, the lines are <u>parallel</u>

  • ║

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = \dfrac{13}{5}

Therefore, the two equations are <u>neither</u> parallel nor perpendicular

  • <u>Neither</u>

7. The equations are;

2·x + 6·y = -3

Which gives y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}

12·y = 4·x + 20

Which gives

y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}

m₁ ≠ m₂

m_1 \neq -\dfrac{1}{m_2}

  • <u>Neither</u>

8. 2·x - 5·y = -3

Which gives; y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}

5·x + 27 = 6

x = -\dfrac{21}{5}

  • Therefore, the slopes are not equal, or perpendicular, the correct option is <u>Neither</u>

Learn more here:

brainly.com/question/16732089

6 0
3 years ago
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