Question

What are the chances that no two boys are sitting together for a photograph if there are 5 girls and 2 boys?

Answer 1

The chances that no two boys are sitting together for a photograph if there are 5 girls and 2 boys is $\dfrac{5}{7}$

Explanation:

Given : The number of girls = 5

the number of boys = 2

Total children = 5+2=7

The number of ways to arrange n things (linear)= n!

Then , the total number of ways to arrange 7 children = 7! = 5040

If two boys sit together  , we consider pair of them as one child , then the total number of children = 5+1=6

Then , the number of ways to arrange 6 children = 6!

Number of ways two boys sit together = 2 x (6! ) = 2(720)= 1440

Number of ways such that no two boys are sitting together =Total number of ways to arrange 7 children - Number of ways two boys sit together

=  5040- 1440 = 3600

Now , the probability that no two boys are sitting together for a photograph if there are 5 girls and 2 boys :

$\dfrac{\text{No two boys sit together}}{\text{Total children}}\\\\=\dfrac{3600}{5040}\\\\=\dfrac{5}{7}$

Hence, the chances that no two boys are sitting together for a photograph if there are 5 girls and 2 boys $=\dfrac{5}{7}$

#Learn more :

Please help me and explain full answer if you it........

In how many ways can 6 boys and 5 girls be arranged for a group photograph if the girls are to sit on chairs in a row and the boys are to stand in a row. Behind them.

https://brainly.in/question/2075198

Answer 2

Step-by-step explanation:

➟7+(

5

−3

)

➠\frac{35 - 3}{5}

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answer is

\frac{5}{7}

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hope you have got your answer!!