Question

5 persons entered the lift cabin on the ground floor of a $8$ floor house (Assume ground floor as also one floor). Assume that each of them independently and with equal probability can leave the cabin at any floor beginning from the first. Find the probability for all the 5 persons to leave the cabin at different floors.

• A. $\frac{7!}{2!\times 7^5}$
• B. $\frac{7!}{2!\times 5^7}$
• C. $\frac{7!}{3!\times 7^5}$
• D. None of these

Answer 1

The correct option is A $\frac{7!}{2!\times 7^5}$

1st person can leave at any of seven floors in $7$ ways

2nd person can leave at any of seven floors in $6$ ways

3rd person can leave at any of seven floors in $5$ ways

4th person can leave at any of seven floors in $4$ ways

5th person can leave at any of seven floors in $3$ ways

Total favourable ways $=7\times 6\times 5\times 4\times 3=2520=\frac{7!}{2!}$

Total possible ways= $7\times 7\times 7\times 7\times 7=7^5$

Probability$=\frac{7!}{(2!\times 7^5)}$