PS - Combinatorics

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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 smae letters used in a different order constitute a different code, how many different stocks is it possible to uniquely deisgnate with these codes?

A) 2,951
B) 8,125
C) 15,600
D) 16,302
E) 18,278

15 sec method its gotta be C..

first code 26 ways..second code 26P2 and 3rd 26P3 now..we 26P2 gives us 25*26..ie there has to be a zero in the unit digit of our ans choice..only C has it..

Ok I think I got it, an initial oversight on my part.

26 + 26^2 + 26^3

you are right..

26 for the first one..26*26 for the second one and 26*26*26 for the 3rd one

total 26+26*26+26*26^3

e it is

That was my initial thoughts and when adding it all together you get something close to answer D (they're or because it can either be 1,2 or 3 letters)

so it would be 26 + 26P2 + 26P3
but order matters and you can have the same letters
so for 3 letters it should be 26x26x26
and for 2 letters 26x26

I get E

I sometimes get confused when to use Permutation and when not to..here i felt order matters..so use permutation..

i guess..there was a simpler way of doing it too..

but anyone here anyone good insight on permutations?

Here you were right in choosing to use permutations.
You can think of permuations as labelling and order
how many different ways can I label the objects and does order matter?

So lets say you have the word GMAT, how many 4 letter combinations of the word can you make?
Each letter has a label of itself G M A T
There are 4 ways to pick the 1st letter
3 ways to pick the 2nd letter
2 to pick the 3rd
1 to pick the 4th
so you have 4x3x2x1 = 4! ways
or another way of saying it is 4P4 = 4!/(4-4)! = 4!/0! where 0! = 1

Now if you have a situation where the labels are the same, so we have the word GMMAAT
2 M's are the same and 2 A's are the same
start with how many ways you can pick all the letters
6!
but 2 labels of M are the same and 2 labels of A are the same
they can both be arranged 2! ways each
so these don't count.
6!/(2!2!)

As an extrapolation from the above,
Situations when to use the combination formula and not permutation is when there can only be 2 labels
so if John Mary Joe Bob and George need to form a team of 2
there will only be 2 labels "on the team" "off the team"
so 5C2
or 5! ways to arrange them all
divided by (2! ways to arrange the people "on the team" because they have the same label)
and also divided by (3!, the ways to arrange the people "off the team" because they have the same label)
so you have 5!/(2!3!) which = 5C2

I hope this makes sense

As pertaining to this particular question
I was a little off in saying the Permutation formula could be used.
You cannot replace the object when the permutation formula is used.
But it is derived from the labels.

So you have GMAT again
and you have to make a 2 letter word
order matters
all the letters are different
4P2 right?
4! ways to rearrange the letters over 2! ways that the "not in the word" letters can be arranged.
"not in the word" is a label in themselves

For this question you can replace the eggs in the basket
so you just have to think of it in terms of choices you have for each letter.
so for 3 letter you have 26x26x26

Hi Glen,

Here's a sure fire way, I haven't thrown eggs away in the last ten years.

I keep a Komodo Dragon under the sink in the kitchen, I let him sniff one of the bad batch, if he turns his nose up at it then it's probably off. If he eats it then is't okay to use the rest. If you've only got one left then you have to be careful, he might bite your fingers off as you taunt him with the egg and he tries to bite it.

Kind of confused

1 - 26
2 - 26*25 - The second letter can't be same as the first letter hence it's not 26*26
3- 26*26*25

Am I right?

they all can be the same
1. 26
2. 26*26
3. 26*26*26
26+676+17576=18278

can i go this way?

26C1
26C2
26C3

26 + 26^2 + 26^3 = 18278

E