Question
A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the $10^{\text{th}\text{~}}$ floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to :
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Solution
Sol. [Key Concept: Split the $9$ people into a group of $4$ and a group of $5$ ($\frac{9!}{4!5!}$), pick $2$ distinct floors from the $8$ allowed, and arrange them ($\times 2$).]
Form group of 4 and 5 person $\frac{9!}{4!5!} = 126$
choose two floor out of 8 floor $\ ^{8}C_{2} = 28$
Total ways in which 4 & 5 persons can leave at different floor $126 \times 28 \times 2 = 7056$
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