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
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:
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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
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there should only be 3 out of the 60 that are in alphabetical order.
they would be:
abc
bcd
cde
no others are possible.
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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.
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