Find the number of permutations of letters `a ,b ,c ,d ,e ,f,g` taken all together if neither `beg` nor `cad` pattern appear.

Explanation

The total number of permutations without any restriction is 7!
`n(U)=7!=5040`
Let n(A) be the number of permutations in which 'beg' pattern always appears
`"b e g a c d f"`
i.e., `n(A)=5!=120`
and let `n(B)` be the number of permutations in which 'cad'
c a d b e f g
i.e., `n(B)=5!=120`
Now, `n(A capB)=` number of permutations inwhich 'beg' and 'cad' pattern appear
b e g c a d f
i.e., `n(A capB)=3!=6`
Hence, the number of permutations in which 'beg' annd 'cad' patterns do not appears is `n(A'capB')`
or `n(A'capB')=n(U)-n(AcupB)`
`=n(U)-[n(A)+n(B)-n(A capB)]`
`=5040-120-120+6=4806`