Question

9 persons enter a lift from the ground floor of a building which stops in $10$ floors (excluding ground floor), if it is known that persons will leave the lift in groups of $2,3,4$ in different floors. In how many ways this can happen?

Answer 1

Hint: Since there are a total nine-person entering into a lift on the ground floor of the building, and that lift will stop at the endpoint of floor ten but the restriction is the ground floor will not be calculated in this problem. So, we need to find the number of ways to find the person who will leave the lift in the group of two, three and four on the different floors. We use a combination formula to solve this problem.
Formula used: ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$(combination)

Complete step-by-step solution:
Let there are a total of nine persons and also three different floors of the exit points, hence by the combination formula we get ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$ here n is the total person that is nine and three is the number of ways to exit; hence we get ${}^9{c_3} = \dfrac{{9!}}{{3!(9 - 3)!}}$ this can be simplified into ${}^9{c_3} = \dfrac{{9 \times 8 \times 7}}{{3 \times 2}}$ and further solving we get ${}^9{c_3} = 84$ ways. Since the first group two persons exit first so ${}^9{c_2}$ ways, and the three-person exit next in ${}^7{c_3}$ ways, and similarly, for last four persons, we have ${}^4{c_4}$ ways.
Floors that can be exited in $3!$ways, hence we need to multiply everything to get the resultant for the required problem. Thus, we get $84 \times 6 \times {}^9{c_2} \times {}^7{c_3} \times {}^4{c_4} = 635040$. Hence there is $635040$ a number of ways that three groups can exit the lift.

Note: Since the given question asks us to find the number of ways so we used the formula of combination as above. If the question is about the number of arrangements, then we must need to use the formula of permutation which is ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$(n is the total number and r is the number of ways to arrange).