Answer to Question #125749 in Statistics and Probability for sanat

Question

eight boys and three girls are to sit in row for a photograph.find the probability that no two girls are together

Answer 1

We are seeking the situations of the form

$(x_1\text{ boys})(\text{a girl})(x_2\text{ boys})(\text{a girl})(x_3\text{ boys})(\text{a girl})(x_4\text{ boys})$,

where $x_2,x_3>0.$ The number of such situations is the number of sequences of nonnegative integers $x_1,x_2,x_3,x_4$ such that $x_1+x_2+x_3+x_4=8$ and $x_2,x_3>0$ times the number of permutations between girls and boys, i.e. $8!\cdot 3!$. The number of those sequences is the number of sequences of positive integers $y_1,y_2,y_3,y_4$ such that $y_1+y_2+y_3+y_4=10$ with the bijection $(x_1,x_2,x_3,x_4)\mapsto(x_1+1,x_2,x_3,x_4+1)=(y_1,y_2,y_3,y_4)$, and we indeed have $y_1+y_2+y_3+y_4=(x_1+1)+x_2+x_3+(x_4+1)=10$. It's bijective with taking a line of $10$ balls and choosing $3$ borders between them with the bijection

$(y_1,y_2,y_3,y_4)\mapsto (\text{borders after the balls }y_1,y_1+y_2,y_1+y_2+y_3).$

We choose $3$ borders from $9$ places between balls, which gives $\binom93$ such sequences. Thus, the probability that no two girls sit near each other equals $\binom93\cdot 8!\cdot 3!/11!=28/55$, where we divide by $11!$ since it's the number of permutations of $8+3=11$ boys and girls.