Probability Ques 19
In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, $3$ in the front and $4$ at the back? How many seating arrangements are possible if $3 $girls should sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of $3$ girls sitting together in a back row on adjacent seats?
$(1996,5\ M)$
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Answer:
Correct Answer: 19.$(\frac{1}{91})$
Solution:
Formula:
- We have $14$ seats in two vans and there are $9$ boys and $3$ girls. The number of ways of arranging $12$ people on $14$ seats without restriction is
$ { }^{14} P _{12}=\frac{14 !}{(14-12) !}=\frac{14 !}{2 !}=7(13 !) $
Now, the number of ways of choosing back seats is $2$ . and the number of ways of arranging $3$ girls on adjacent seats is $2(3 !)$ and the number of ways of arranging $9$ boys on the remaining $11$ seats is ${ }^{11} P _9$ ways.
Therefore, the required number of permutations
$ =2 \cdot(2 \cdot 3 !) \cdot{ }^{11} P _9=\frac{4 \cdot 3 ! \cdot 11 !}{2 !}=12 ! $
Hence, the probability of the required event occurring
$ =\frac{12 !}{7 \cdot 13 !}=\frac{1}{91} $
BETA
