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Maths Question 7 – JEE-MAIN 2026

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A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the 10th floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to :

First, determine the total number of floors where passengers can exit, considering the given restrictions.

Step 1: Determine Valid Exit Floors✦ Active

The building has a ground floor and 10 more floors, making floors 0,1,2,...,10. The lift goes up to the 10th floor. The lift does not stop at the 1st and 2nd floors. Therefore, the available floors for exit are 3rd,4th,...,10th floors. The total number of available floors is 10−3+1=8 floors.

Step 2: Select Persons and Assign Floors○ Expand

First, choose 4 persons out of 9. This can be done in 9C4 ways. The remaining 5 persons form the second group. Next, choose a floor for the group of 4 persons from the 8 available floors. This can be done in 8 ways. Finally, choose a *different* floor for the group of 5 persons from the remaining 7 available floors. This can be done in 7 ways.

9C4=9!4!5!=9×8×7×64×3×2×1=126
💡 Teacher's Secret Hint

Remember that the two groups must exit at *different* floors, which implies an ordered selection of floors.

Step 3: Calculate Total Number of Ways○ Expand

The total number of ways is the product of the ways to choose the persons and the ways to assign them to distinct floors.

Total ways=Ways to choose 4 persons×Ways to choose 1st floor×Ways to choose 2nd floor
Total ways=126×8×7=126×56=7056

Thus, the number of ways is 7056.

💡 Teacher's Secret Hint

Ensure all conditions, especially the 'different floor' condition, are correctly incorporated into the calculation.

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