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5 persons entered a lift cabin in the cellar of a 7 -floor building apart from cellar. If each of them independently and with equal probability can leave the cabin at any floor out of the 7 floors beginning with the first, then the probability of all the 5 persons leaving the cabii at different floors is
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Verified Answer
The correct answer is:
$\frac{360}{2401}$
Total floors $=7$
Total number of ways to exit $=7^5$
If 5 persons leave at 5 different floors then number of ways $=$ arranging 5 people in 7 ways $={ }^7 \mathrm{P}_5$
Required probability $=\frac{{ }^7 \mathrm{P}_5}{7^5}$
$$
=\frac{7 !}{2 ! .7^5}=\frac{360}{2401}
$$
Total number of ways to exit $=7^5$
If 5 persons leave at 5 different floors then number of ways $=$ arranging 5 people in 7 ways $={ }^7 \mathrm{P}_5$
Required probability $=\frac{{ }^7 \mathrm{P}_5}{7^5}$
$$
=\frac{7 !}{2 ! .7^5}=\frac{360}{2401}
$$
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