Question

The number of permutation of letters $a,b,c,d,e,f,g$ so that neither the pattern $beg$ nor $cad$ appears is

• A. $\frac{7!}{3!3!}$
• B. $\frac{7!}{3!3!31}$
• C. $4806$
• D. None of these

Answer 1

Answer: (C)

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 $=7!-234=4806$.