The Number of ways in which five different books to be distributed among 3 persons so that each person gets at least one book, is equal to the number of ways in which?

Options

A. (a)5 persons are allotted 3 different residential flats so that each person is allotted at most one flat and no two persons are allotted the same flat
B. (b) Number of parallelograms (some of which may be overlapping) formed by one set of 6 parallel lines and other set of 5 parallel lines that goes in other direction
C. (c) 5 different toys are to be distributed among 3 children, so that each child gets at least one toy
D. (d) 3 mathematics professors are assigned five different lectures to be delivered, so that each professor gets at least one lecture

Answer

To solve the problem of distributing five different books among three persons such that each person gets at least one book, we can use the principle of inclusion-exclusion. Let's break down the steps:

### Step 1: Identify the Total Ways to Distribute Books
First, we need to calculate the total number of ways to distribute the five different books to three persons without any restrictions. Each book can go to any of the three persons. Therefore, the total number of distributions is given by:
\[ 3^5 \]
This is because for each of the 5 books, there are 3 choices (one for each person).

### Step 2: Subtract the Cases Where At Least One Person Gets No Book
Now, we need to subtract the cases where at least one person does not receive a book. We can use the principle of inclusion-exclusion for this.

1. **Choose 1 person to be excluded**: There are \( \binom{3}{1} = 3 \) ways to choose one person who will not receive any books. The remaining two persons can receive the books. The number of ways to distribute the books among the remaining two persons is \( 2^5 \).

Therefore, the total ways for this case is:
\[ 3 \times 2^5 = 3 \times 32 = 96 \]

### Step 3: Add Back the Cases Where Two Persons Get No Books
Next, we need to add back the cases where two persons do not receive any books (since we subtracted these cases twice).

2. **Choose 2 persons to be excluded**: There are \( \binom{3}{2} = 3 \) ways to choose two persons who will not receive any books. The remaining one person will receive all the books. There is only 1 way to distribute all books to one person.

Therefore, the total ways for this case is:
\[ 3 \times 1 = 3 \]

### Step 4: Calculate the Final Count
Now we can combine these results using the inclusion-exclusion principle:
\[ \text{Total ways} = 3^5 - \text{(ways with at least 1 person getting no book)} + \text{(ways with at least 2 persons getting no book)} \]
\[ \text{Total ways} = 243 - 96 + 3 = 150 \]

### Conclusion
Thus, the number of ways to distribute five different books among three persons such that each person gets at least one book is 150.