Question

A lift is going up to the \(10^{\text{th}}\) floor. Number of ways in which \(3\) people can exit the lift at {three different floors, if the lift will not stop at I$^{\text{st}}$, II$^{\text{nd}}$ and III$^{\text{rd}}$ floors, is: }

• A. \(210\)
• B. \(343\)
• C. \(720\)
• D. \(205\)

Hint:

When {people are distinct} and {floors are distinct}, use {permutations} rather than combinations.

Answer 1

The Correct Option is A

Step 1: Identify Available Floors
The lift goes up to the \(10^{\text{th}}\) floor but does not stop at: \[ 1^{\text{st}},\ 2^{\text{nd}},\ 3^{\text{rd}} \] So, possible stopping floors are: \[ 4, 5, 6, 7, 8, 9, 10 \] Total available floors: \[ 7 \]
Step 2: Condition of the Problem


There are \(3\) distinct people.
Each person must get down at a different floor
.

Step 3: Count the Number of Ways
Number of ways to assign \(3\) different floors to \(3\) distinct people from \(7\) floors: \[ {}^{7}P_{3} = 7 \times 6 \times 5 = 210 \]
Final Answer:
\[ \boxed{210} \]