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Maths Question 9 – JEE-MAIN 2026

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A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is:

This problem involves distributing distinct items (books) into distinct containers (bags) with the constraint that no container remains empty.

Step 1: Calculate Total Distribution Ways✦ Active

We have 4 different books and 3 different bags. Each book can be placed in any of the 3 bags. Therefore, the total number of ways to distribute the 4 books into the 3 bags without any restrictions is 34.

Total ways=34=81
Step 2: Apply Principle of Inclusion-Exclusion for Empty Bags○ Expand

We need to subtract the cases where one or more bags are empty. Using the Principle of Inclusion-Exclusion:

1. **At least one bag is empty:** Choose 1 bag to be empty in (31) ways. The 4 books must go into the remaining 2 bags. This can be done in 24 ways. So, (31)×24=3×16=48 ways.

2. **At least two bags are empty:** Choose 2 bags to be empty in (32) ways. The 4 books must go into the remaining 1 bag. This can be done in 14 ways. So, (32)×14=3×1=3 ways.

3. **All three bags are empty:** Choose 3 bags to be empty in (33) ways. The 4 books must go into the remaining 0 bags. This is 04=0 ways.

The number of ways where at least one bag is empty is 48−3+0=45.

💡 Teacher's Secret Hint

Remember to alternate signs in the Inclusion-Exclusion Principle.

Step 3: Calculate Ways with No Empty Bags○ Expand

The number of ways to put the books in the bags so that no bag is empty is the total number of ways minus the number of ways where at least one bag is empty.

Ways (no empty bags)=Total ways−Ways (at least one bag empty)
=81−45=36
💡 Teacher's Secret Hint

This problem can also be solved using Stirling numbers of the second kind: k!S(n,k), where n=4,k=3. 3!S(4,3)=6×6=36.

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