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: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(2)

Bag 1: 2

Bag 2: 1

Bag 3: 1

Number of ways = $(\frac{(4!)}{2! {(1!)}^{2} . 2!}) \times 3!$

$= \frac{24 \times 6}{2 \times 2} = 36$

Solution

Q.
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: [2026]

1 18

2 36

3 39

4 72

Ans.
(2)

Bag 1: 2

Bag 2: 1

Bag 3: 1

Number of ways = $(\frac{(4!)}{2! {(1!)}^{2} . 2!}) \times 3!$

$= \frac{24 \times 6}{2 \times 2} = 36$

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