The question asks for the number of ways three distinct persons can exit a lift at three different floors, with specific floors excluded.
The lift can go up to the 10th floor. However, it does not stop at the 1st, 2nd, and 3rd floors. Therefore, the floors available for exiting are 4, 5, 6, 7, 8, 9, and 10.
The total number of available floors is $10 - 3 = 7$.
There are 3 distinct persons, and they must exit at 3 different floors chosen from the 7 available floors. Since the persons are distinct, the order in which they choose the floors matters.
This scenario requires calculating permutations, specifically the number of ways to arrange 3 items (floors) chosen from a set of 7 items (available floors).
The number of permutations of selecting $k$ items from a set of $n$ items is given by the formula $P(n, k) = \frac{n!}{(n-k)!}$.
Here, $n = 7$ (the number of available floors) and $k = 3$ (the number of persons).
The number of ways is calculated as:
$P(7, 3) = \frac{7!}{(7-3)!} = \frac{7!}{4!}$
$P(7, 3) = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}$
$P(7, 3) = 7 \times 6 \times 5$
$P(7, 3) = 210$
There are 210 distinct ways for the three persons to exit the lift at three different floors from the available options.
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.