Question

The number of permutation from the letter A to G so that neither the set BEG nor CAD appears is

• A. $\frac{7!}{(3!{)}^2}$
• B. $3!\times (3{)}^2\times 89$
• C. $\frac{7!}{(3!{)}^3}$
• D. None of these

Answer 1

The correct option is B $3!\times (3{)}^2\times 89$

Total number of permutations $=7!$
Let L denote that 'BEG' occurs and M denote that CAD occurs .

Number of permutation with $L=5!$

Number of permutation with $M=5!$

Number of permutations when both L and M occur $(L\cap M)=3!$
$\therefore n(L\cup M)=5!+5!-3!=234$
$\therefore$ Required number of ways $=7!-234=4806=3!\times 3^2\times 89$
Hence (b) is correct answer.