Three persons enter in a lift at the ground floor. The lift will go upto 10th floor. The number of w

Q: Why can't we just use $^7C_3$?

$^7C_3$ only selects the floors, but it ignores the fact that the people are distinct. If Person A gets off at floor 4 and Person B at floor 5, that is a different scenario than Person B at floor 4 and Person A at floor 5. We must account for this arrangement, which is why we use permutations.

Q: Why do we subtract 3 from 10?

The problem explicitly states the lift does not stop at the first, second, and third floors. These are 'forbidden' zones. We must remove them from the total set of 10 floors, leaving us with $10 - 3 = 7$ valid floors to choose from.

Q: Is the order of the people important?

Absolutely. In combinatorics, whenever we are dealing with distinct items (like people) being placed into distinct slots (like floors), the order of assignment matters. Therefore, we use the permutation formula $^nP_r$ rather than the combination formula $^nC_r$.

Question

Three persons enter in a lift at the ground floor. The lift will go upto $10^{th}$ floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the lobby of a grand building with ten floors. Three people walk into the lift at the ground floor. Their journey is about to begin, and they can go anywhere up to the tenth floor.

The problem states that the lift will not stop at the first, second, and third floors. These are forbidden zones. We must cross them out.

By removing these three floors, we are left with a set of available floors: $\{4, 5, 6, 7, 8, 9, 10\}$. Counting these, we find that our total number of available floors is $n = 10 - 3 = 7$. This is our playground.

### The Logic of Choice: Permutation vs. Combination

Now, consider the three people. They are distinct individuals.

If Person A exits at floor 4 and Person B at floor 5, it is a completely different outcome than if Person B exits at floor 4 and Person A at floor 5. Because the order of exit matters, we are dealing with a permutation problem.

We need to select and arrange 3 distinct floors out of the 7 available ones. This is where the permutation formula $^nP_r$ becomes our best friend. We have $n = 7$ and $r = 3$.

### The Mathematical Execution

Let us set up the equation. We need to calculate $^7P_3$. The general formula for permutations is:

$$^nP_r = \frac{n!}{(n-r)!}$$

Substituting our values, we get:

$$^7P_3 = \frac{7!}{(7-3)!}$$

Simplifying the denominator, we have $7 - 3 = 4$, so the expression becomes:

$$\frac{7!}{4!}$$

Instead of calculating the full value of $7!$, we can expand it until we hit $4!$:

$$7! = 7 	imes 6 	imes 5 	imes 4!$$

Now, our expression is:

$$\frac{7 	imes 6 	imes 5 	imes 4!}{4!}$$

The beauty of this step lies in the cancellation. The $4!$ in the numerator and the $4!$ in the denominator cancel out perfectly, leaving us with a straightforward multiplication: $7 	imes 6 	imes 5$.

Calculating this, $7 	imes 6 = 42$, and $42 	imes 5 = 210$.

There are exactly **210** ways for these three people to exit the lift. You have successfully navigated the constraints and applied the correct combinatorial logic.

Enter Numerical Value:

Visualized Solution

Visualizing the Scenario

Identifying Forbidden Floors

Counting Available Floors

The Condition for Exiting

Applying Permutation Logic

Setting Up the Equation

Substituting the Values

Simplifying the Denominator

Expanding the Factorial

Canceling Terms

Final Calculation

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Logic of Choice

The Mathematical Execution

Similar Questions

In a group of 3 girls and 4 boys, there are two boys $B_{1}$ and $B_{2}$ . The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_{1}$ and $B_{2}$ are not adjacent to each other, is:

A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to__________

$m$ men and $n$ women are to be seated in a row so that no two women sit together. If $m > n$ , then show that the number of ways in which they can be seated is $\frac{m ! ( m + 1 )!}{( m - n + 1 )!}$

Let the digits $a, b, c$ be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is ____.