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Goryan [66]
3 years ago
10

A group of 10 people need to form a line. The line will consist of exactly 8 of the people. Person X and Person Y have to be eit

her fifth or sixth in line. How many different orders are possible?
Mathematics
1 answer:
skad [1K]3 years ago
6 0

Answer: N = 40,320

Step-by-step explanation:

To arrange 10 people to a line of 8 people.

The number of ways is n = 10P8.

But since the condition given states that 2 people must stay in two particular slots.

The number of possible arrangements becomes;

N = number of ways of arranging the remaining people × number of ways of arranging the other 2.

N = 8P6 × 2P2

N = 8!/2! × 2!

N = 20,160 × 2

N = 40,320 ways

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Answer:

The probability of obtaining a sequence which is neither strictly increasing nor decreasing in 0.76.

Step-by-step explanation:

We have 10 possible outcomes on a single throw.

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Let y be the number of strictly decreasing arrangements.

Let z be the number of outcomes that are neither strictly decreasing nor strictly increasing

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If we look at a strictly increasing arrangement from the other/opposite side, it will look like a strictly decreasing arrangement.

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So, the probability is = \frac{760}{1000} = 0.76

Therefore, the probability of obtaining a sequence which is neither strictly increasing nor decreasing in 0.76.

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