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

The correct option is C

$4806$

The explanation for the correct option.

The given letters are $a,b,c,d,e,f,g$.

Number of letters$=7$

The number of permutations of letters $a,b,c,d,e,f,g$$=7!$

The letters with pattern $beg$ are $a,c,d,f,\left(beg\right)$

The number of permutations of letters with pattern $beg$$=5!$

The letters with pattern $cad$ are $b,e,f,g,\left(cad\right)$

The number of permutations of letters with pattern $cad$$=5!$

The letters with pattern both $cad$ and $beg$ are $f,\left(cad\right),\left(beg\right)$.

The number of permutations of both $cad$ and $beg$ come together in the arrangement$=3!$

The number of permutations in which neither ‘$beg$’ nor $cad$ appear is computed as,

$$\begin{gathered} =7!-(5!+5!-3!) \\ =(7\times 6\times 5\times 4\times 3\times 2\times 1)-(120+120-6) \\ =5040-234 \\ =4806 \end{gathered}$$

Thus, the number of permutations in which neither ‘$beg$’ nor $cad$ appear is $4806$.

Hence, the option $C$ is correct.