Question 549349: Three letters are randomly selected without replacement one at a time from {a,b,c,d----Z}.What is the probability that they are selected in alphabetical order. Express your answer as a common fraction. This was a question in 2010 state math counts
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there are 26 letters in the alphabet.
probability of the first letter being in alphabetical order is 24 out of 26.
last letter possible is letter x because y and z would come after that.
y is no good because only 1 letter left after that in alphabetical.
z is no good because no letters left after that in alphabetical order.
probability of the second letter being in alphabetical order is 1 out of the remaining 25.
probability of the last letter being in alphabeticak order is 1 out of the remaining 24.
total probability is therefore 24/26 * 1/25 * 1/24
to see how this works, assume only 5 letters.
letters are a, b, c, d, e.
probability here, if the formula is correct, would be 3/5 * 1/4 * 1/3 = 3/60
let's see if that's correct.
first letter could be a, b, or c
if a, then b and c have to follow in that order.
if b, then c and d have to follow in that order.
if c, then d and e have to follow in that order.
total possible combinations possible combinations of letters possible when order is important are:
5P3 which comes out to be 5! / 2! which comes out to be (5*4*3*2*1) / (1*2) which comes out to be 60 possible permutations.
they are:
-----
abc acb bac bca cab cba
abd adb bad bda dab dba
abe aeb bae bea eab eba
acd adc cad cda dac dca
ace aec cae cea eac eca
ade aed dae dea ead eda
bcd bdc cbd cdb dbc dcb
bce bec cbe ceb ebc ecb
bde bed dbe deb ebd edb
cde ced dce dec ecd edc
-----
there should only be 3 out of the 60 that are in alphabetical order.
they would be:
abc
bcd
cde
no others are possible.
-----
since the same formula was applied to the 26 letters in the alphabet, then the answer for the alphabet should be as good.
let's see if that's true.
the probability for the alphabet was 24/26 * 1/25 * 1/24 which became (24*1*1) / (26*25*24) = 24 / 15600
the number of possible permutations of 3 out of 26 is 26! / 23! = (26*25*24*23!) / 23! = (26*25*24) = 15600.
out of these, the total number of possible sequences of letters in alphabetical order are:
abc
bcd
cde
def
efg
fgh
ghi
hij
ijk
jkl
klm
lmn
mno
nop
opq
pqr
qrs
rst
stu
tuv
uvw
vwx
wxy
xyz
yz
z
you can see that only 24 out of the 26 letters of the alphabet have the possibility of being in alphabetical order 3 at a time.
the total possible number of permutations of 3 out of 26 is given by the permutation formula of n! / (n-x)! where n is the total number of possible choices to pick from (26) and x is the number you pick at a time (3).
26! / (26-3)! becomes 26! / 23! which is the number of 15600 we arrived at earlier.
Answer by ikleyn(53762)  (Show Source):
You can put this solution on YOUR website! .
Three letters are randomly selected without replacement one at a time from {a,b,c,d----Z}.
What is the probability that they are selected in alphabetical order.
Express your answer as a common fraction. This was a question in 2010 state math counts
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The set {a,b,c,d---Z) contains 26 + 26 = 52 letters.
Let's consider all possible triples of letters from this set without repeating.
Every triple has 6 incarnations (6 possible permutations).
Of these 6 permutations, ONLY ONE is in alphabetical order.
So, the probability under the problem's question is  .
It is the ANSWER to the problem.
Good problem and a nice solution.
I would say even more: this problem and the solution in some sense
revolve your mind and teach you " to think in the other way ".
The reasoning can be constructed/re-told in other way,
giving " another solution ".
The number of all ordered triples with no repeating is
P(52,3) = 52*81*50. (1)
The number of all unordered triples (=combinations) is
C(52,3) =  . (2)
Each single combination represents 6 permutations from P(52,3), which we count
as one ordered triple in the alphabetical order.
Therefore, the seeking probability is the number of all different alphabetically ordered triples C(52,3)
to the number of all possible ordered triples P(52,3).
From formulas (1) and (2), it is clear that this ratio is
P =  = .
This completes the second solution. We get the same answer.
-------------------------------------------
The solution by @Theo is 73 lines long and they all go out the target.
His solution has many words, but is incorrect, and even has no right idea.
////////////////////////////////////////////////////////
Now I want to present and to discuss another interesting fact.
Today, Oct.17, 2025, I submitted this problem to Google AI Overview,
which works now as a universal free of charge solver to Math problems in the Internet.
Below is the Google AI Overview solution to this problem.
I copy-pasted it today, Oct.17, 2025, 2:56 PM
- - - - - - - - AI Overview START - - - - - - -
The probability is 1/26*1/25*1/24, which is equal to 1/15600.
The total number of ways to select three letters in order is found
by multiplying the number of options for each selection: 26*25*24 = 15600.
Only one of these ways is in alphabetical order.
1. Total possible ordered selections: Since letters are selected without
replacement, there are 26 choices for the first letter, 25 for the second,
and 24 for the third. The total number of ordered combinations is .
2. Favorable outcome: There is only one way to select three letters that are in
alphabetical order (e.g., {a, b, c} in that specific order).
3. Calculate the probability: The probability is the number of favorable
outcomes divided by the total number of possible outcomes: [1, 3, 4, 5, 6]
AI responses may include mistakes.
[1] https://www.youtube.com/watch?v=7xu8kKVzT-w
[2] https://brainly.com/question/30269108
[3] https://gre.myprepclub.com/forum/three-letters-are-randomly-selected-one-at-a-time-without-replacement-23170.html
[4] https://www.gauthmath.com/solution/1760404381135877/2-A-computer-randomly-selects-a-letter-from-the-alphabet-a-How-many-different-ou
[5] https://study.com/skill/practice/calculating-probabilities-of-draws-without-replacement-questions.html
[6] https://www.cliffsnotes.com/study-notes/24757581
The link to this Google AI solution
https://www.google.com/search?q=Three+letters+are+randomly+selected+without+replacement+one+at+a+time+from+%7Ba%2Cb%2Cc%2Cd----Z%7D.+What+is+the+probability+that+they+are+selected+in+alphabetical+order.&rlz=1C1CHBF_enUS1071US1071&oq=Three+letters+are+randomly+selected+without+replacement+one+at+a+time+from+%7Ba%2Cb%2Cc%2Cd----Z%7D.+What+is+the+probability+that+they+are+selected+in+alphabetical+order.&gs_lcrp=EgZjaHJvbWUyBggAEEUYOTIGCAEQRRg8MgYIAhBFGDzSAQkxODU4ajBqMTWoAgiwAgHxBZKTgCYYjWrL&sourceid=chrome&ie=UTF-8
- - - - - - - - AI Overview END - - - - - - -
Below is my (@ikleyn) comment.
This AI solution is FATALLY WRONG. It demonstrates complete and absolute misunderstanding of the subject.
So, it shows the level of authority of this solver.
What is good that at the end of the solution, this AI solver honestly warns, that his output may include mistakes.
This is exactly the case.
It also characterizes the level of the solutions in the Internet, on which Google AI Overview
bases his conclusions. They also are at the zero level of understanding the subject.
Naturally, I reported to the Google AI Overview about this their mistake through their feedback system,
and provided the link to this my post.
Hope, it will help to the AI to become a bit smarter.
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
In conclusion, couple more words to a reader.
A standard approach is this formal approach which is my solution #2.
Very often, it is the first move to many students to start thinking this way.
There is nothing bad in it, but it requires some level of maturity.
Only those can go this way to the end, who are able "to break through the wall
with his/her forehead", that is, with his/her mind.
In other words, it requires certain level of training.
Therefore, if you have difficulties moving forward this way - - - try another way,
as I did in my solution #1.
May be, you intuition and/or your common sense will help you in this way.
It is what I call " to think differently ".
|
|
|
