Q.9. A person has three different bags and four different books. The number of ways, in which he can

Q: What is the correct answer for JEE Main Mathematics Q9?

The correct answer is B. Let the three different bags be $B_1, B_2, B_3$ and four different books be $b_1, b_2, b_3, b_4$. Each book can be put into any of the 3 bags. So, for 4 books, there are $3^4$ total ways to put the books into the bags. Total ways = $3^4 = 81$. Now, we need to find the number of ways such that no bag is empty. We can use the Principle of Inclusion-Exclusion. Let $N$ be the total number of ways = $

Q: Q.9. 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

The correct answer is B. Let the three different bags be $B_1, B_2, B_3$ and four different books be $b_1, b_2, b_3, b_4$. Each book can be put into any of the 3 bags. So, for 4 books, there are $3^4$ total ways to put the books into the bags. Total ways = $3^4 = 81$. Now, we need to find the number of ways such that no bag is empty. We can use the Principle of Inclusion-Exclusion. Let $N$ be the total number of ways = $

Q: What are the options for this JEE Main Mathematics question?

A. 18 · B. 36 · C. 39 · D. 72

Question

Q.9. 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

FirstInTest · Accepted Answer

Let the three different bags be $B_1, B_2, B_3$ and four different books be $b_1, b_2, b_3, b_4$.
Each book can be put into any of the 3 bags. So, for 4 books, there are $3^4$ total ways to put the books into the bags.
Total ways = $3^4 = 81$.

Now, we need to find the number of ways such that no bag is empty. We can use the Principle of Inclusion-Exclusion.
Let $N$ be the total number of ways = $3^4 = 81$.
Let $A_i$ be the property that bag $B_i$ is empty.
We want to find $N - |A_1 \cup A_2 \cup A_3|$.

$|A_1|$: Bag $B_1$ is empty. The 4 books can be placed in $B_2$ or $B_3$. So $2^4$ ways. Similarly, $|A_2| = 2^4$ and $|A_3| = 2^4$.
$\sum |A_i| = 3 	imes 2^4 = 3 	imes 16 = 48$.

$|A_1 \cap A_2 \cap A_3|$: All three bags are empty. This is 0 ways (as books must be placed).

Number of ways = $N - \sum |A_i| + \sum |A_i \cap A_j| - |A_1 \cap A_2 \cap A_3|$
$= 81 - 48 + 3 - 0 = 36$.

Alternatively, this is equivalent to distributing 4 distinct items into 3 distinct boxes such that no box is empty, which is $3! S(4,3)$, where $S(n,k)$ is the Stirling number of the second kind.
$S(4,3) = \frac{1}{3!} (3^4 - 3 \cdot 2^4 + 3 \cdot 1^4 - 0) = \frac{1}{6} (81 - 48 + 3) = \frac{36}{6} = 6$.
Number of ways = $3! 	imes 6 = 6 	imes 6 = 36$.

The solution uses the formula for surjective mappings or distribution of distinct items into distinct boxes: $k! S(n,k)$.
Here, $n=4$ (books), $k=3$ (bags).
Number of ways = $3! 	imes S(4,3) = 3! 	imes (\frac{1}{3!} (3^4 - ^{3}C_1 2^4 + ^{3}C_2 1^4)) = 3^4 - 3 \cdot 2^4 + 3 \cdot 1^4 = 81 - 3 \cdot 16 + 3 \cdot 1 = 81 - 48 + 3 = 36$.

Answer

18

Answer

36

Answer

39

Answer

72

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