JEE Mathematics 2026 Question with Solution

Given: Three persons enter a lift at the ground floor. The lift goes up to the **$10$**th floor and does not stop at **$1$**st, **$2$**nd and **$3$**rd floors.

Find: The number of ways in which the three distinct persons can exit at three different floors.

Allowed exit floors are:

$$4,5,6,7,8,9,10$$

So, the total number of available floors is $7$.

The three persons must get down at three different floors, so first choose $3$ distinct floors from $7$ floors:

$${}^7C_3 = 35$$

Now assign the $3$ distinct persons to these $3$ chosen floors:

$$3! = 6$$

Hence, total number of ways is:

$$\text{Total ways} = {}^7C_3 \times 3! = 35 \times 6 = 210$$

Therefore, the required number of ways is $210$.

Given: The lift can stop only at floors $4$ to $10$.

Find: The number of distinct assignments of three persons to three different exit floors.

First select the floors:

$$\binom{7}{3} = 35$$

Then distribute the three persons among those selected floors:

$$3! = 6$$

Thus,

$$35 \times 6 = 210$$

Therefore, the correct answer is $210$.