If n things are arranged in a row, the number of way in which they can be arranged so that non occupies its original position is `n!(1-(1)/(1!)+(1)/(2!)-(1)/(3!)+* * * + (-1)^(n)(1)/(n!))`
Five boys are to be seated in a row. The number of ways in which 3 boys are not seated in the place specfied to them is

Options

A. `""^(5)P_(2)`
B. 10
C. 40
D. 25

Answer

To solve the problem of arranging 5 boys in a row such that 3 specific boys do not sit in their specified places, we can follow these steps:

### Step 1: Understand the Problem
We have 5 boys and we need to ensure that 3 of them do not sit in their assigned positions. This means that 2 boys will sit in their specified places.

### Step 2: Select the Boys
We need to select 3 boys from the 5 boys. The number of ways to choose 3 boys from 5 is given by the combination formula \( C(n, r) \), where \( n \) is the total number of boys and \( r \) is the number of boys to choose.

\[ C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 3: Arrange the Selected Boys
Now, we need to arrange the 3 selected boys such that none of them sits in their original position. This is a classic derangement problem, denoted as \( !n \) (the number of derangements of \( n \) items).

For 3 boys, the formula for derangement is:

\[ !n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{(-1)^n}{n!}\right) \]

Calculating for \( n = 3 \):

\[ !3 = 3! \left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) = 6 \left(0 + \frac{1}{2} - \frac{1}{6}\right) \]

Calculating the terms inside the parentheses:

\[ = 6 \left(\frac{3}{6} - \frac{1}{6}\right) = 6 \left(\frac{2}{6}\right) = 6 \cdot \frac{1}{3} = 2 \]

### Step 4: Calculate Total Arrangements
Now, we multiply the number of ways to choose the boys by the number of derangements:

\[ \text{Total arrangements} = C(5, 3) \times !3 = 10 \times 2 = 20 \]

Thus, the total number of ways in which 3 boys can be seated such that none of them occupies their original position is **20**.

### Final Answer
The number of ways in which 3 boys are not seated in the places specified to them is **20**.

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