Question

The number of permutations 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. $\frac{7!}{2!3!3!}$
• C. $4806$
• D. None of these

Answer 1

Answer: (C)

$4806$

The total number of permutations $=7!$

Let A be the property that $beg$ occurs and B be the property that $cad$ occurs.

Number of permutations with A$=5!$=that of with B and $n(A\cap B)$$=3!$

Thus $n(A\cap B)=5!+5!-3!=234$

Thus required number $=7!-234=4806$