# Table and chairs

Four people should sit at a table in front of a row of 7 chairs. What is the probability that there will be no empty chair between them if people choose their place completely at random?

Correct result:

p =  0.114

#### Solution:

$C_{{ 4}}(7)=\dbinom{ 7}{ 4}=\dfrac{ 7! }{ 4!(7-4)!}=\dfrac{ 7 \cdot 6 \cdot 5 } { 3 \cdot 2 \cdot 1 }=35 \ \\ \ \\ n_{1}={ { 7 } \choose 4 }=35 \ \\ n_{2}=4 \ \\ n_{3}=0 \ \\ p=n_{2}/n_{1}=4/35=\dfrac{ 4 }{ 35 }=0.114$

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