Question

In an eleventh storey building 9 people enter a lift cabin. It is known that they will leave the lift in groups of 2, 3 and 4 at different residential storeys. Find the number of ways in which they can get down.

Answer 1

Hint: First we divide the 9 people into three groups of 2, 3 and 4 members respectively. For that we select 2 members from 9 people, then 3 members from the remaining 7 people and then we select the remaining 4 members. Then by multiplying them we will get the number of ways of dividing 9 people into 3 groups of 2, 3 and 4 members. Now we have 3 groups and we have to find the number of ways of dropping them in 10 floors, that is equal to the number of ways of selecting 3 floors from 9 floors and by multiplying them we get our answer.
Complete step by step answer:
Let us divide the 9 people into 3 groups of 2, 3 and 4 members.
Number of ways of selection of 2 people is \[{}^{9}{{C}_{2}}\].
Now the remaining number of people are 7, now let us select 3 people from the 7, then the number of ways of selection of 3 people is \[{}^{7}{{C}_{3}}\].
Now the number of people remaining is 4 and we have to make a group of 4, then the number of ways of selecting 4 people from the remaining 4 people is \[{}^{4}{{C}_{4}}\].
Total number of ways of dividing 9 people into 3 groups of 2, 3 and 4 members is \[{}^{9}{{C}_{2}}\times {}^{7}{{C}_{3}}\times {}^{4}{{C}_{4}}\].
\[=\dfrac{9!}{(9-2)!\times 2!}\times \dfrac{7!}{(7-3)!\times 3!}\times \dfrac{4!}{(4-4)!\times 4!}=\dfrac{9\times 8}{2}\times \dfrac{7\times 6\times 5}{3\times 2\times 1}\times \dfrac{4\times 3\times 2\times 1}{4\times 3\times 2\times 1}=9\times 4\times 7\times 5=1260\]
Now we have 3 groups, so they can leave the lift in 3! = 6 ways.
The number of floors where the person exits = 3.
So, total ways = ${}^{10}{{C}_{3}}=\dfrac{10\times 9\times 8}{3\times 2\times 1}=120$.
Now let us multiply the values we got to get the number of ways to drop 3 groups of 2, 3 and 4 members into 3 different floors.
\[=1260\times 6\times 120=907200\]
Hence, the required number of ways \[=907200\].

Note:
The possibilities for making mistakes in this type of problems are, one may make a mistake by considering the total number of ways of that the 9 people will leave the lift in group of 2, 3 and 4 as $\dfrac{9!}{2!3!4!}$.