<s>[INST] probability of $3$ girls sitting together in back row of adjacent seat in vans In how many ways $9$ boys and $3$ girls can be selected in two vans,each having numbered seats,$3$ in the front and $4$ at the back? How many sitting arrangements are possible $3$ girls should sit together in a back row of adjacent seat? Now if all the sitting arrangements are equal likely, what is the probability of $3$ girls sitting together in back row of adjacent seat
what i try
we have $14$ seats in $2$ vans and there are $9$ boys and $3$ girls. The number of ways of arranging $12$ people on $14$ seats without ristriction is $\displaystyle \binom{14}{12}\times 12=7\times 13!$
how do i arrange $3$ girls sit together in back seat help me please
 [/INST] We have $9$ boys $+$ $3$ girls $=12$ people
Total seats in one van $=3+4=7$
Total seats in $2$ vans $=7\times2=14$
So, we have to arrange $12$ people in $14$ seats.
Number of ways $= \ ^{14}P_{12}=\dfrac{14!}{(14-12)!}=7(13!)$
Now the number of ways to choose back seats is $2$ and we can arrange girls in adjacent seats in $2(3!)$ ways.
Number of ways of arranging $9$ bos in the remaining $11$ seats is $^{11}P_9$ ways.
Therefore, the total number of ways are $2\times2\times3!\times^{11}P_9=12!$
Therefore, the required probability is $\dfrac{12!}{7\times13!}=\dfrac{1}{91}$
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