In how many ways can the letters of the English alphabet be arranged so that there are exactly ten letters between a and z?

Solution:

$\begin{aligned} &\text { There are } 26 \text { letters in the English alphabet. If we separate the group }(a \text {, some } 10 \\ &\text { letters, z), we will be left with } 14 \text { more letters. } \\ &\text { These } 15 \text { objects ( } 1 \text { group }+14 \text { letters) can be arranged among themselves in } \\ &15 ! \text { ways. } \\ &\text { Since either a or z can be at the beginning or the end of the group of } 12 \text { letters }(a \text {, } \\ &\text { some } 10 \text { letters, b), the number of possible arrangements of the group will be } 2 \times \\ &\left({ }^{1} P_{1} \times{ }^{10} P_{10} \times{ }^{1} P_{1}\right)=2 \times 10 ! \text {. } \\ &\text { Also, each group of } 10 \text { letters can be selected from the remaining } 24 \text { letters } \\ &\left(\text { except a and b) in }{ }^{24} C_{10}\right. \text { ways. } \\ &\text { Required total number of ways }=\left(2 \times 10 ! \times{ }^{24} C_{10}\right) \times 15 ! \\ \end{aligned}$