The number of permutations of the letters a_1 | vedclass

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(B) Let $S$ be the set of all permutations of the $5$ letters,so $|S| = 5! = 120$. Let $P_1$ be the property that $a_1$ is in the first position,and $

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$78$

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(B) Let $S$ be the set of all permutations of the $5$ letters,so $|S| = 5! = 120$. Let $P_1$ be the property that $a_1$ is in the first position,and $P_2$ be the property that $a_2$ is in the second position. We want to find the number of permutations that satisfy neither $P_1$ nor $P_2$,which is given by $|S| - |P_1 \cup P_2| = |S| - (|P_1| + |P_2| - |P_1 \cap P_2|)$. $|P_1|$ is the number of permutations where $a_1$ is fixed at the first position,which is $4! = 24$. $|P_2|$ is the number of permutations where $a_2$ is fixed at the second position,which is $4! = 24$. $|P_1 \cap P_2|$ is the number of permutations where both $a_1$ is at the first position and $a_2$ is at the second position,which is $3! = 6$. Thus,$|P_1 \cup P_2| = 24 + 24 - 6 = 42$. The number of permutations where $a_1$ is not in the first position and $a_2$ is not in the second position is $120 - 42 = 78$.

The number of permutations of the letters $a_1, a_2, a_3, a_4, a_5$ in which the first letter $a_1$ does not occupy the first position and the second letter $a_2$ does not occupy the second position is:

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