The correct match of the following is :
`{:(,"Column - I",,"Column - II"),((i),""^(n)C_(r ),(a),n!),((ii),""^(n)C_(0),(b),1),((iii),""^(n)P_(r ),(c ),r!.""^(n)C_(r )),((iv),""^(n)P_(n),(d),""^(n)C_(n-r)):}`

Options

A. (i)-(d), (ii)-(b), (iii)-(c ), (iv)-(a)
B. (i)-(a), (ii)-(b), (iii)-(c ), (iv)-(d)
C. (i)-(b), (ii)-(c ), (iii)-(d), (iv)-(a)
D. (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c )

Answer

To solve the problem, we need to match the items from Column I with their corresponding items in Column II based on the properties of combinations and permutations.

### Step-by-Step Solution:

1. **Understanding \( nCr \)**:
- The formula for combinations is given by:
\[ nCr = \frac{n!}{r!(n-r)!} \]
- A key property of combinations is that:
\[ nCr = nC(n-r) \]
- Thus, for \( i \) (which corresponds to \( nCr \)), the match is \( d \) (which corresponds to \( nC(n-r) \)).

2. **Understanding \( nC0 \)**:
- The value of \( nC0 \) is calculated as:
\[ nC0 = \frac{n!}{0!(n-0)!} = \frac{n!}{1 \cdot n!} = 1 \]
- Therefore, for \( ii \) (which corresponds to \( nC0 \)), the match is \( b \) (which corresponds to \( 1 \)).

3. **Understanding \( nPr \)**:
- The formula for permutations is given by:
\[ nPr = \frac{n!}{(n-r)!} \]
- This can also be expressed in terms of combinations:
\[ nPr = r! \cdot nCr \]
- Therefore, for \( iii \) (which corresponds to \( nPr \)), the match is \( c \) (which corresponds to \( r! \cdot nCr \)).

4. **Understanding \( nPn \)**:
- The value of \( nPn \) is calculated as:
\[ nPn = \frac{n!}{(n-n)!} = \frac{n!}{0!} = n! \]
- Since \( nPn \) is simply \( n! \), for \( iv \) (which corresponds to \( nPn \)), the match is \( a \) (which corresponds to \( n! \)).

### Final Matches:
- \( i \) matches with \( d \) ( \( nCr = nC(n-r) \) )
- \( ii \) matches with \( b \) ( \( nC0 = 1 \) )
- \( iii \) matches with \( c \) ( \( nPr = r! \cdot nCr \) )
- \( iv \) matches with \( a \) ( \( nPn = n! \) )

### Summary of Matches:
- \( (i, d) \)
- \( (ii, b) \)
- \( (iii, c) \)
- \( (iv, a) \)