Probability 1 Question 19
27. 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: 27. 0.62
Solution:
- 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
$$ { }^{14} P _{12}=\frac{14 !}{2 !}=7(13 !) $$
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 ways
$$ =2 \cdot(2 \cdot 3 !) \cdot{ }^{11} P _9=\frac{4 \cdot 3 ! 11 !}{2 !}=12 ! $$
Hence, the probability of the required event
$$ =\frac{12 !}{7 \cdot 13 !}=\frac{1}{91} $$
