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3 years ago

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A family is taking a picture, with everyone in the family standing in a row. In the family, there are two sets of twins. In how

many ways can the 9 people in the family be arranged such that nobody is sitting next to their twin?

1 answer:

Wittaler [7]3 years ago

5 0

Answer:

376320.

Step-by-step explanation:

Given:

There are two sets of twins in a family.

Total family member = 9

Solution:

Let family members = AA BB C D E F G

<u>AA and BB are two set of twins .</u>

First of we will find the arrangements of 5 family member ( 9 - 4 = 5 )

No of ways 5 family member can seat = $^{5}P_{5}$

Now, we will find the arrangements of BB.

As given that nobody is sitting next to their twin, means A twins cannot seat together.

Therefore, they can seat in these blank places given below:

......B.......B......C......... D...... E........ F........ G.......... in $^{8}P_{2}$

Similarly, B twins cannot seat together.

Therefore, they can seat in these blank places given below:

......A.......A......C......... D...... E........ F........ G.......... in $^{8}P_{2}$

Total number of arrangements of all family members = $^{5}P_{5}$ $\times$ $^{8}P_{2}$ $\times$ $^{8}P_{2}$

$=\frac{5!}{5-5!} \times\frac{8!}{8-2!} \times\frac{8!}{8-2!}\\=\frac{5!}{0!}\times\frac{8!}{6!} \times\frac{8!}{6!}\\\\=5!\times\frac{8\times7\times6!}{6!} \frac{8\times7\times6!}{6!}$

$=5\times4\times3\times2\times1\times56\times56\\=376320$

Therefore, Total number of arrangements of all family members is 376320.

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The distance around a triangle is 46 ft the length 12 what is the width

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<em><u>Solution:</u></em>

Given that,

We have to simplify

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We can simplify the above expression by combining the like terms

Like terms are terms that has same variable with same exponent and same or different coefficient

From given,

$w^2 + 7w^2 + 8w -5+9-2w^2$

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3 years ago

John runs a computer software store. Yesterday he counted 123 people who walked by the store, 56 of whom came into the store. Of

vaieri [72.5K]

Answer:

a) There is a 45.53% probability that a person who walks by the store will enter the store.

b) There is a 41.07% probability that a person who walks into the store will buy something.

c) There is a 18.70% probability that a person who walks by the store will come in and buy something.

d) There is a 58.93% probability that a person who comes into the store will buy nothing.

Step-by-step explanation:

This a probability problem.

The probability formula is given by:

$P = \frac{D}{T}$

In which P is the probability, D is the number of desired outcomes and T is the number of total outcomes.

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123 people walked by the store.

56 people came into the store.

23 bought something in the store.

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So

$P = \frac{D}{T} = \frac{56}{123} = 0.4553$

There is a 45.53% probability that a person who walks by the store will enter the store.

(b) Estimate the probability that a person who walks into the store will buy something.

56 people came into the store and 23 bought something, so $T = 56, D = 23$ .

So

$P = \frac{D}{T} = \frac{23}{56} = 0.4107$

There is a 41.07% probability that a person who walks into the store will buy something.

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$P = \frac{D}{T} = \frac{23}{123} = 0.1870$

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$P = \frac{D}{T} = \frac{33}{56} = 0.5893$

There is a 58.93% probability that a person who comes into the store will buy nothing.

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Answer:

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