How many ways can 4 books be distributed in 3 bags with no bag empty?

Using inclusion-exclusion: 3^4 − C(3,1)·2^4 + C(3,2)·1^4 = 81−48+3 = 36.

What formula is used for surjective distributions?

∑_{i=0}^{k} (−1)^i · C(k,i) · (k−i)^n for distributing n distinct objects into k distinct non-empty boxes.

3 different bags and 4 different books — number of ways to put books so no bag is empty

JEERANKERS

MathP&C

QMCQPermutations & Combinations

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:

A) $18$ B) $36$ C) $39$ D) $72$

✅ Correct Answer

B) 36

✓

Solution

Total distributions (no restriction)

Each of 4 books can go into any of 3 bags independently:

$$\text{Total} = 3^4 = 81$$

Apply Inclusion-Exclusion to remove cases with empty bags

At least 1 specific bag empty: $\binom{3}{1}\cdot2^4=3\times16=48$

At least 2 specific bags empty: $\binom{3}{2}\cdot1^4=3\times1=3$

All 3 bags empty: $\binom{3}{3}\cdot0^4=0$

Apply the formula

$$\text{No bag empty} = 3^4-\binom{3}{1}\cdot2^4+\binom{3}{2}\cdot1^4-\binom{3}{3}\cdot0^4$$ $$=81-48+3-0=\boxed{36}$$

Verify using partition method

To distribute 4 distinct books among 3 distinct bags with no empty bag, the partition type must be $(2,1,1)$.

Choose 2 books for one bag: $\binom{4}{2}=6$ ways.
Assign this pair to one of 3 bags: $3$ ways.
Remaining 2 books go to remaining 2 bags (1 each): $2!=2$ ways.

Total $=6\times3\times2=36$ ✓

📘 Surjective Distribution Formula

To distribute $n$ distinct objects into $k$ distinct boxes with no box empty (surjection), use: $\sum_{i=0}^{k}(-1)^i\binom{k}{i}(k-i)^n$. For $n=4,k=3$: $81-48+3=36$. Alternatively, $k!\cdot S(n,k)$ where $S(n,k)$ is the Stirling number of the second kind.

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Important Concepts

Inclusion-Exclusion PrincipleRemove cases where at least one bag is empty: $|A_1\cup A_2\cup A_3|=\sum|A_i|-\sum|A_i\cap A_j|+|A_1\cap A_2\cap A_3|=48-3+0=45$. Answer $=81-45=36$.

Partition Type (2,1,1)For 4 books in 3 bags with no empty bag, exactly one bag gets 2 books. Ways: choose 2 books ($\binom{4}{2}=6$) × assign to a bag ($3$) × distribute remaining ($2!$) $=36$.

Stirling Number Approach$S(4,3)=6$ (Stirling number of second kind = ways to partition 4 elements into 3 non-empty unlabelled groups). Multiply by $3!=6$ to label the bags: $6\times6=36$.

Why 3^4 for total?Without restriction, each of 4 books independently chooses 1 of 3 bags: $3\times3\times3\times3=81$. This counts ALL distributions including those with empty bags.

FAQs

What does 'no bag empty' mean combinatorially?

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It means the distribution is a surjective (onto) function from the set of books to the set of bags — every bag must receive at least one book.

Why subtract C(3,1)·2^4?

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$\binom{3}{1}$ chooses which specific bag is empty; $2^4$ counts distributions of all 4 books among the remaining 2 bags. By inclusion-exclusion, this is subtracted.

What is the Stirling number S(4,3)?

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$S(4,3)=6$ counts the ways to partition 4 elements into exactly 3 non-empty unlabelled subsets. For labelled bags, multiply by $3!=6$ to get $36$.

If the bags were identical, what would the answer be?

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Then we only count partitions (not arrangements), so $S(4,3)=6$ ways.

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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