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Q.
The number of permutation of letters $a, b, c, d, e, f, g $ so that neither the pattern $beg$ nor $cad$ appears is
Permutations and Combinations
Solution:
Total number of permutations $= 7!$
Let $A$ be the permutations that $beg$ occurs.
and $B$ be the permutations that $cad$ occurs.
Number of permutations with $A$ (or $B$) $= 5!$
and $n (A \cap B) = 3! $
$\therefore n(A \cup B) = 5! + 5! - 3! = 234$
$\therefore $ Required number $= 7! - 234 = 4806$.