My attempt:
the answer of first question is: $$A_{26}^5=\frac{26!}{21!}=26*25*24*23*22=7893600$$
But i got stuck in the second question but i think the answer is $A_{6}^2 A_{20}^3$ is it true ?
2026-03-28 06:48:17.1774680497
• How many ”special” words are in $S$
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Your answer is correct. There are $6$ choices for the second letter, and then $5$ for fourth, giving you $\ A_6^2\ $ for the number of ways those two letters can be chosen. Independently of that, there are $20$ ways of choosing the first letter, then $19$ of choosing the third and $18$ of choosing the fifth, giving you $\ A_{20}^3\ $ ways of choosing those three letters. Therefore, the total number of ways of choosing the five letters to satisfy the second condition is your answer of $\ A_6^2A_{20}^3\ $.