Question

Among the 5 married couples, if the names of 5 men are matched with the names of their wives randomly, then the probability that no man is matched with the name of his own wife is ?

• A. \(\tfrac{9}{20}\)
• B. \(\tfrac{1}{5}\)
• C. \(\tfrac{11}{30}\)
• D. \(\tfrac{17}{60}\)

Hint:

For “no item in its original place,” use the derangement count \( !n \).

The probability is \( \frac{!n}{n!} \).

Answer 1

The Correct Option is C

Step 1: Standard derangement formula.
The number of ways to permute 5 items such that none is in its original place (a “derangement”) is denoted by \(!5\). A known formula: \[ !n = n!\sum_{k=0}^{n}\frac{(-1)^k}{k!}. \] Hence \[ !5 = 5!\Bigl(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\Bigr) =120\Bigl(1-1+\tfrac12-\tfrac16+\tfrac{1}{24}-\tfrac{1}{120}\Bigr). \] \[ =120\Bigl(\tfrac12-\tfrac16+\tfrac{1}{24}-\tfrac{1}{120}\Bigr) =120\Bigl(\tfrac{60-20+5-1}{120}\Bigr) =120\cdot\frac{44}{120} =44. \] So there are 44 derangements for 5 items. 

Step 2: Probability that no man is matched with his wife.
Total ways to match 5 men's names with 5 wives' names is \(5!\! =120\). The favorable ways (derangements) is 44. Thus \[ P(\text{no man matched to own wife}) = \frac{44}{120} = \frac{11}{30}. \]