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What is the probability of the event that z precedes both a and b when we randomly select a permutation of the 26 lowercase letters of the English alphabet?

Currently, my thoughts are:

P(25,25)x24 There are p(25,25) ways to get certain letters of the alphabet

24 ways to place z, it can't be in the last 2 spots since it has to precede a
and b.

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You have ruled out some ways to place $z$, but it is obviously possible to place $z$ somewhere in the first 24 spots without it preceding $a$ and $b$. So there is no hope that your strategy can be correct, since you are counting combinations that do not belong. –  Erick Wong Dec 1 '12 at 3:03
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2 Answers 2

up vote 4 down vote accepted

There are $6$ permutations of our set of $3$ letters. In $2$ of them, z precedes both a and b, so the probability is $\dfrac{2}{6}$.

The locations of the other letters is irrelevant. If this is not obvious, note that given any specific locations for the other $23$ letters, all of the $3!$ orders of our target letters are equally likely, and $2$ of these orders, z, a, b and z, b, a satisfy our condition.

Remark: One can also do it the hard(er) way. There are $26!$ permutations of the letters, all equally likely. Now we count the permutations in which $z$ precedes $a$ and $b$. There are $\dbinom{26}{3}$ ways to decide on the set of locations that will be occupied by the set of letters a, b, c. Once we have done that, there are $2$ ways to arrange a, b, and z in these locations so that z is first. And then there are $23!$ ways to arrange the remaining letters. So our probability is $$\frac{\binom{26}{3}(2)(23!)}{26!}.$$ This simplifies to $\dfrac{1}{3}$.

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One of the three letters must come first, and each has an equal chance! –  hardmath Dec 1 '12 at 1:10
    
I liked your showing the "hard(er) way" gives the same answer. I suspect the OP learns most when the agreement between the shortcut and cumbersome approaches is laid bare. –  hardmath Dec 2 '12 at 1:12
    
What am I missing here... C(26, 3) = 2600, 2600*2 = 5200, 5200/26! is obviously not 1/3. Is C(26, 3) not the same as (26 3)? –  Colton Nov 24 '13 at 4:23
    
@Colton: Thanks, you are not missing anything. The $23!$ was in the text but not in the displayed formula. Fixed. –  André Nicolas Nov 24 '13 at 5:41
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By symmetry, each of a, b, z is equally likely to be the one that precedes the other two. So the probability is $1/3$.

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Better!${}{}{}{}{}{}{}$ –  André Nicolas Dec 1 '12 at 1:14
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