JEE Advanced 2020 Paper 2Question 1
In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ____
Permutation and Combination
Answer$1080$
Solution
Room sizes must be $2,2,1,1$. Choose the two double rooms: $\binom42=6$; distribute persons: $\frac{6!}{2!\,2!}=180$. Total $6\cdot180=1080$.