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vladimir1956 [14]
2 years ago
7

There are five different routes that a commuter can take from her home to the office. In how many ways can she make a round trip

if she uses a different route coming than going?
Mathematics
1 answer:
Lynna [10]2 years ago
4 0

Answer: 120 ways

Step-by-step explanation:

Let's label the routes a, b, c, d, and e. The goal is to come up with as many unique permutations as possible.

1 route: a=1.               one factorial (1!) =1*1=1

2 routes: ab, ba=2.      2!=1*2=2

3 routes: abc, acb, bac, bca, cab, cba=6.      3!=1*2*3=6

4 routes: 4!=1*2*3*4=24

5 routes: 5!=1*2*3*4*5=120 ways

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the only prime number in between those numbers is 43.

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Andre45 [30]

Answer:

  • A) \displaystyle \frac{1}{77}.
  • B) \displaystyle \frac{12}{77}.
  • C) \displaystyle \frac{4}{11}.

Step-by-step explanation:

All marbles here are identical. Also, the question isn't concerned about the order in which the marbles are drawn. Thus, all calculations here shall be combinations rather than permutations.

<h3>A)</h3>

How many ways to choose three out of six identical red marbles without replacement?

\displaystyle _6C_3 = c(6, 3) = {6\choose 3} = 20.

Note that these three expressions are equivalent. They all represent the number of ways to choose 3 out of 6 identical items without replacement.

How many ways to choose three out of all the 6 + 10 + 6 = 22 marbles?

\displaystyle _{22} C_{3} = 1540.

The probability of choosing three red marbles out of these 22 marbles will be:

\displaystyle \frac{\text{Number of ways for choosing three out of six red marbles}}{\text{Number of ways to choose three out of 22 marbles}} = \frac{20}{1540} = \frac{1}{77}.

<h3>B)</h3>

How many ways to choose two out of six identical red marbles without replacement?

\displaystyle _6 C_2 = 15.

How many ways to choose one out of 10 + 6 = 16 non-red marbles?

_{16} C_1=16.

Choosing two red marbles does not influence the number of ways of choosing a non-red marble. Both event happen and are independent of each other. Apply the product rule to find the number of ways of choosing two red marbles and one non-red marble out of the pile of 22.

_6 C_2 \cdot _{16} C_1= 240.

Probability:

\displaystyle \frac{240}{1540} = \frac{12}{77}.

Double check that the order doesn't matter here.

<h3>C)</h3>

None of the marbles are red. In other words, all three marbles are chosen out of a pile of 10 + 6 = 16 white and blue marbles. Number of ways to do so:

_{16} C_{3} = 560.

Probability:

\displaystyle \frac{560}{1540}= \frac{4}{11}.

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