Mathematics, 27.10.2019 05:43 dpazmembreno

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

a. the first 13 letters of the permutation are in alphabetical order.

b. a is the first letter of the permutation and z is the last letter.

c. a and z are next to each other in the permutation.

d. a and b are not next to each other in the permutation.

e. a and z are separated by at least 23 letters in the permutation. f ) z precedes both a and b in the permutation?

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Answer from: MAKYERAHWILLIAMS09

There are 26! (about 403*10^24) possible permutations of the 26 letters.

a. We're allowing permutations matching the pattern

a b c ... k l m ...b c d ... l m n ...c d e ... m n o ...

and so on.

Quick note: If the permutations starts with "o", then we could allow cyclic permutations, such as

o p q ... y z a ...z a b ... j k l ...

to be in alphabetical order, but not

o p q ... y z b ...z a b ... j k m ...

The first letter determines the next 12, and we have 26 choices for the first letter. The remaining 13 positions can take any of the remaining 13 letters. So there are 26*13! such permutations, and the probability is

$\dfrac{26\cdot13!}{26!}=\dfrac{13!}{25!}\approx4.01\times10^{-16}$

On the other hand, if we want to be stricter about what alphabetical order entails and prohibit permutations starting with "o" or any other letter of the alphabet following "o", then the first position has 14 possible choices ("a" through "n"). Then there are 14*13! = 14! such permutations, and the probability of this occurring is

$\dfrac{14!}{26!}=\approx2.16\times10^{-16}$

b. We're fixing the first and last positions, leaving 24 positions to be filled by any of the remaining 24 letters. There are 24! such permutations, so the probability of this happening is

$\dfrac{24!}{26!}=\dfrac1{26\cdot25}=\dfrac1{650}\approx0.0015$

c. I'll assume here that permutations like

a z ..._ z a ...

are okay, but not

z ... a

The number of permutations containing "a z" is the same as the number of permutations containing "z a". There are 25 valid positions for "a" to take and precede "z". The remaining 24 positions can take on any other letter. So the number of such permutations is 25*24! = 25!. We double this to count the same permutations with "a" and "z" swapped. So the probability is

$\dfrac{2\cdot25!}{26!}=\dfrac1{13}\approx0.0769$

d. Use the same strategy as in part (c). There are 2*25! permutations containing "a b" or "b a". We want the permutations that don't have this subsequence, so we subtract this number from the total number of permutations. So the probability is

$\dfrac{26!-2\cdot25!}{26!}=1-\dfrac1{13}=\dfrac{12}{13}\approx0.9231$

e. We first need to count how many ways we can have a distance of 23 between "a" and "z". Luckily there are only two cases:

if "a" is at 1st position, then "z" can be at either 25th or 26thif "a" is at 2nd position, then "z" must be at 26th

all 3 of which involve 24! ways of permuting the remaining 24 letters, so we multiply by 2. We have to multiply by 2 again to account for swapping "a" and "z". So the probability is

$\dfrac{2\cdot3\cdot24!}{26!}=\dfrac3{325}\approx0.0092$

f. "z" has 24 possible positions because at least 2 of the positions after the occurrence of "z" must include "a" and "b". There's a simple pattern to the number of permutations here:

if "z" is at 1st, then there are 25! ways of permuting the remaining lettersif "z" is at 2nd, then there are 23 choices of letter for 1st position, and 24! ways of permuting the restif "z" is at 3rd, then there are 23*22 choices for positions 1-2, and 23! ways of permuting the restif "z" is at 4th, then there are 23*22*21 choices for positions 1-3, and 22! ways of permuting the rest

and so on, so that there are

$\displaystyle\sum_{n=0}^{23}\frac{23!}{(23-n)!}(25-n)!$