In an eleven 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. Number of ways in which they can get down is:

Question

In an eleven 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. Number of ways in which they can get down is: (A) ((12) !/2 !(3 !)2 ⋅ 4 !) (B) (10 !/4) (C) (10 !/4 !) (D) ((12) !/(3 !)2 ⋅ 4 !)

Tardigrade Admin · Accepted Answer

9 people can be divided into groups of 2,3 and 4 in (9 !/2 ! 3 ! 4 !) ways. Now each group can be distributed in 10 C 3 × 3 ! ways ⇒ (9 !/2 ! 3 ! 4 !) 10 ⋅ 9 ⋅ 8=(10 !/4) Note that in an eleven storeyed building there will be 10 floors.

Q. In an eleven 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. Number of ways in which they can get down is:

$\frac{( 12 )!}{2 ! ( 3 ! ) ^{2} \cdot 4 !}$

$\frac{10 !}{4}$

$\frac{10 !}{4 !}$

$\frac{( 12 )!}{( 3 ! ) ^{2} \cdot 4 !}$

$9$ people can be divided into groups of $2, 3$ and $4$ in $\frac{9 !}{2 ! 3 ! 4 !}$ ways.
Now each group can be distributed in $^{10} C_{3} \times 3!$ ways $\Rightarrow \frac{9 !}{2 ! 3 ! 4 !} 10 \cdot 9 \cdot 8 = \frac{10 !}{4}$
Note that in an eleven storeyed building there will be 10 floors.

Q. In an eleven 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. Number of ways in which they can get down is:

2!(3!)2⋅4!(12)!​

410!​

4!10!​

(3!)2⋅4!(12)!​