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Mathematics · JEE | PERMUTATIONS AND COMBINATIONS

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Match List-I with List-II: List-I List-II A \( {^n\text{P}_r} \) I \( \frac{n!}{r!(n-r)!} \) B \( {^n\text{C}_r} \) II \( \frac{n!}{(n-r)!} \) C \( {^n\text{C}_0} \) III \( n! \) D \( {^n\text{P}_n} \) IV \( 1 \) Choose the correct answer from the options given below:

Match List-I with List-II:

List-I List-II
A \( {^n\text{P}_r} \) I \( \frac{n!}{r!(n-r)!} \)
B \( {^n\text{C}_r} \) II \( \frac{n!}{(n-r)!} \)
C \( {^n\text{C}_0} \) III \( n! \)
D \( {^n\text{P}_n} \) IV \( 1 \)

Choose the correct answer from the options given below:

  1. A
    A-(II), B-(I), C-(IV), D-(III)
  2. B
    A-(I), B-(II), C-(III), D-(IV)
  3. C
    A-(III), B-(IV), C-(I), D-(II)
  4. D
    A-(II), B-(III), C-(IV), D-(I)
View Answer & Solution

Correct answer · Option A

A-(II), B-(I), C-(IV), D-(III)

Explanation

Matching formulas: \( {^n\text{P}_r} = \frac{n!}{(n-r)!} \), \( {^n\text{C}_r} = \frac{n!}{r!(n-r)!} \), \( {^n\text{C}_0} = 1 \), and \( {^n\text{P}_n} = n! \).
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