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What is the probability of these events when we randomly select a permutation of the $26 $ lowercase letters of the English alphabet?

$a) $The permutation consists of the letters in reverse alphabetic order.

$b)$ $z$ is the first letter of the permutation.

$c) z$ precedes $a$ in the permutation.

$d) a$ immediately precedes $z$ in the permutation.

e) a immediately precedes m which immediately precedes z in permutation

please help me with number e

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    for (d), consider $az$ as one letter, now you have $24$ letters and this special letter. So $25!$ ways. – 2017-01-08
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    yeah thanks it works just fine – 2017-01-08
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    yeah can i treat number e the same way u dealt with number d – 2017-01-08
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    yes $amz$ is one special letter. – 2017-01-08
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    worked as well ..... one last question what if he said they are in their original places i highly doubt it would be the same because it is no longer a string they are seperate yeah ? – 2017-01-08
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    I'm assuming you mean $a.......m.........z$. Then think of it as letting the "other" $23$ guests take their seats in any way they want (so how many ways???) Once they have taken the seat then the three special guests $a$, $m$ and $z$ will walk-in to take the first, the thirteenth and the twenty sixth seat respectively. – 2017-01-08

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a) There are $26!$ permutations and only $1$ is like this, so $1/26!$.
b) Each letter is equally likely to be first, so $1/26$.
c) Pair up each permutation with its reverse. Exactly one of each pair has z before a, and one has z after a, so exactly half the permutations have z before a.
d)

If z has to be immediately before a, we are looking for permutations of the 25 objects $b,c,...,y$ and $(az)$. So there are 25! permutations of this form, out of $26!$ total, giving a probability of $1/26$.

e) This can be dealt with in the same way as d).

Now you have permutations of 24 objects, one being (amz), so the answer is $24!/26!=1/650$.

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    Not that it affects the answer but $a$ is preceding $z$. – 2017-01-08