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For a party 7 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together, is |
$\frac{2}{7}$ $\frac{2}{9}$ $\frac{1}{9}$ $\frac{4}{9}$ |
$\frac{2}{9}$ |
The total number of ways in which 7 guests and the couple can sit in a row is 9!. Number of ways in which the husband and his wife sit together is 2! × 8! ∴ Required probability $=\frac{2! × 8!}{9!}=\frac{2}{9}$ |
