Solution
The total possible number of ways is
[tex]\begin{gathered} P(11,4)=\frac{11!}{(11-4)!} \\ \\ P(11,4)=\frac{11!}{7!} \\ \\ P(11,4)=\frac{11\times10\times9\times8\times7!}{7!} \\ \\ P(11,4)=11\times10\times9\times8 \\ \\ P(11,4)=7920 \\ \\ Note:\text{ } \\ P(n,r)=\frac{n!}{(n-r)!},\text{ which is the permutation of n objects selecting r} \end{gathered}[/tex]
The probability of getting a particular four-letter sequence is
[tex]\frac{1}{7920}[/tex]
