A certain stock exchange designates each stock with a one, t

[Guest]

A certain stock exchange designates each stock with a one, two or three letter code , where each letter is selected from the 26 letters of the alphabet. if the letters may be repeated and if the same letters used in a different order constitute a different code , how many different stocks is it possible to uniquely designate with these codes?
a) 2,951
b)8,125
c)15,600
d) 16,302
e) 18,278

OA is E.....18,278.

Please explain this question?

[scrooge]

since repeated letters is ok we take permutation : 1 letter code can be obtained in 26 ways
2 letter can be 26 * 26 ways

3 letter can 26 * 26 * 26

26 + 26^2+26^3 IMO E

[Guest]

Is there a way to solve this problem using permutations? For example, if we are trying to find the # of 2 letter codes using 26 letters can you say:

26! / 2! 24! --> 26 * 25 / 2 = 650.

This is obviously wrong since 26^2 = 675.

What am I doing wrong?

[RonPurewal]

Is there a way to solve this problem using permutations? For example, if we are trying to find the # of 2 letter codes using 26 letters can you say:

26! / 2! 24! --> 26 * 25 / 2 = 650.

This is obviously wrong since 26^2 = 675.

What am I doing wrong?

well, this problem doesn't really involve permutations. in a bona fide permutation, you're not allowed to re-use the letters, whereas this problem permits the re-use of letters.
for example, there's nothing wrong with the stock code LLL, which would be disallowed in an actual permutation.

by the way, 26 x 26 is 676, not 675. i'll assume you just mis-typed that, since two even numbers clearly won't give a product like that.

[RonPurewal]

since repeated letters is ok we take permutation : 1 letter code can be obtained in 26 ways
2 letter can be 26 * 26 ways

3 letter can 26 * 26 * 26

26 + 26^2+26^3 IMO E

correct.

by the way, note that there's no need to actually calculate the values of 26^2 and 26^3. instead, just notice that the units digits of the answer choices are all different, no doubt by design. since this is the case, you can just deal with the units digits only, rather than wasting time multiplying out 676 x 26 to find the perfect cube.

no matter how many 6's you multiply together, you still get a number that ends with a 6. therefore, you're basically just adding together three 6's, which gives 18 --> the number will end with an 8.
answer = (e).

[rahul_bitsp]

Hello Ron.
The question mentions that
"letters may be repeated and if the same letters used in a different order constitute a different code"
but I fail to understand a small issue.
In a 2 code scenario, when we multiple 26 * 26 = 676, are we not including the cases where the first and second symbols are same? I mean don't we count AA and AA (BB or BB) twice ?

I hope you got my question.

Rgds,
Rahul.

[RonPurewal]

In a 2 code scenario, when we multiple 26 * 26 = 676, are we not including the cases where the first and second symbols are same? I mean don't we count AA and AA (BB or BB) twice ?

nope. each distinct "two-letter word" is counted exactly once. in this approach, there's no difference between "AA", which is counted only once, and, say, "DJ", which is also counted only once.
you are correct in that "DJ" and "JD" are counted separately, but that's because they're different "words".

try it yourself, with just A's and B's.
the multiplication gives 2 x 2 = 4.
list them yourself: AA, AB, BA, BB. total 4. the multiplication works.

[sudaif]

if the ORDER DID NOT MATTER, then with the slot method

would it become:

26/1! + 26*26/2! + 26*26*26/3!

?

Thanks

[RonPurewal]

if the ORDER DID NOT MATTER, then with the slot method

would it become:

26/1! + 26*26/2! + 26*26*26/3!

?

Thanks

there's no simple way to do "order doesn't matter" if you allow repetition.
in situations where repetition is allowed, there is actually no known formula for combinations in which order doesn't matter. so if you get a problem like that, your only recourse will be to make a list, and count the possibilities.

it goes without saying that they would only do that with very limited possibilities (i.e., nowhere near the number of outcomes that would happen here).

[pad_bathuu]

I think solution to this problem is...
1.First set of lettters A1,B1, C1.... etc.
Then we have 26 stocks(different stock codes).
2.Second set of letters A2, B2, C2... etc.
Now we have 26+26=52 different letters.
2 letter code out of 52 letters (order matters)
52!/(2!*50!)=1326(different stock codes).
3.Third set of letters A3, B3, C3..... etc
Now we have 26+26+26=78 different letters.
3 letter code out of 78 letters (order matters)
78!/(3!*75!)=76076
The number of total codes for the stock exchange is 76076+1326+26=77428.
This is what I think, but GMAT answer is E.
Please let me know better way to figure out the problem.

[RonPurewal]

Please let me know better way to figure out the problem.

did you read the thread?

the "better way" you're seeking is right here in the thread.

[tanyatomar]

HI Ron,
why cant we do this question like this :

1. 1 letter code=> 26 ways
2. 2 letter code: both letters diff=> 26*25
: both letters same => (26c1 *1)/2!
3. 3 letter code: all letters same: (26c1*1*1)/3!
2 letters same: (26c1*1*25)/2!
all letters different 26*25*24


i did it like this and then dint feel like solving it... :(

[RonPurewal]

HI Ron,
why cant we do this question like this :

1. 1 letter code=> 26 ways
2. 2 letter code: both letters diff=> 26*25
: both letters same => (26c1 *1)/2!
3. 3 letter code: all letters same: (26c1*1*1)/3!
2 letters same: (26c1*1*25)/2!
all letters different 26*25*24


i did it like this and then dint feel like solving it... :(

these aren't right.

whenever possible, you should use normal real-world thinking to evaluate your results.
in this problem, two of your cases are "two-letter code, both same letter" and "three-letter code, all same letter". it should be clear, basically by common sense, that there are exactly 26 possibilities -- the number of letters in the alphabet -- for each of these cases. you are dividing 26 by factorials. not only doesn't that click with real-world intuition, but it actually gives a fractional value when you divide by 3!.

--

if you're going to take this kind of approach, you shouldn't be dividing by any factorials at all, because this is an "order matters" situation. you are talking about codes, so the order of the letters definitely matters.

[tanyatomar]

ohk... thanks a lot for your answer Ron..
i get my mistake..
i was dividing by 3! and 2! because i was taking it like : out of n things if r are of one kind and q are of other kind then the arangements possible are n!/(r!q!)..

basically thinking that if two same kind of things are in a group=> we have to divide by2! ...but in this case it doesnot hold.

in the above example its p n q distinct same kind of things. in the question we are actually picking up the same alphabet so we shud not divide by 3! or 2!..

[RonPurewal]

in the above example its p n q distinct same kind of things. in the question we are actually picking up the same alphabet so we shud not divide by 3! or 2!..

sort of, but that's still really not the issue. the issue is "does order matter, or not?"
here's what i mean:
if the question says "how many three-letter passwords can you make with no letters repeated?" then that's just 26 x 25 x 24, because, when you have a password, the order of the letters matters. (if the letters of a password are in a different order, that's a completely different password.)
on the other hand, if the question is "from a set of 26 tiles labeled with the letters a through z, john selects a set of 3 tiles. in how many ways can john select this set of tiles?" then you have (26 x 25 x 24)/3!, because, in a set of letters, the order is irrelevant. (if the letters in a set are in a different order, it's still the same set.)

so, note that, contrary to what you wrote above, "selecting from the same alphabet" is not the issue. in both of these examples you are selecting from the same alphabet, but one of them requires division by a factorial (because order is immaterial) but the other does not (because order matters).