In how many ways can 5 children be arranged in a line such that (i) two of them, Ram and Shyam, are always together; (ii) two of them,Ram and Shyam, are never together ?

Options

A. a. 12, 42
B. b. 14, 40
C. c. 48, 72
D. d. 28, 80

Answer

To solve the problem, we need to find the number of ways to arrange 5 children under two different conditions:

### Part (i): Ram and Shyam are always together

1. **Consider Ram and Shyam as a single unit**: Since Ram and Shyam must be together, we can treat them as one block or unit. This means we now have 4 units to arrange: (Ram-Shyam), Child 3, Child 4, and Child 5.

2. **Calculate the arrangements of the 4 units**: The number of ways to arrange 4 units is given by \(4!\) (4 factorial).
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

3. **Arrange Ram and Shyam within their block**: Since Ram and Shyam can be arranged in 2 different ways (Ram first or Shyam first), we multiply the arrangements of the units by the arrangements of Ram and Shyam.
\[ 2! = 2 \quad \text{(ways to arrange Ram and Shyam)} \]

4. **Total arrangements when Ram and Shyam are together**:
\[ \text{Total} = 4! \times 2! = 24 \times 2 = 48 \]

### Part (ii): Ram and Shyam are never together

1. **Calculate the total arrangements of all 5 children**: The total number of ways to arrange 5 children is given by \(5!\) (5 factorial).
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

2. **Subtract the arrangements where Ram and Shyam are together**: From the total arrangements, we subtract the arrangements where Ram and Shyam are together (which we calculated in part (i) as 48).
\[ \text{Total arrangements where Ram and Shyam are never together} = 5! - \text{(arrangements where Ram and Shyam are together)} \]
\[ = 120 - 48 = 72 \]

### Final Answers:
- (i) The number of ways to arrange the children such that Ram and Shyam are always together is **48**.
- (ii) The number of ways to arrange the children such that Ram and Shyam are never together is **72**.