fresinha12 wrote:

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?

gmatatouille wrote:

fresinha12 wrote:

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..

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

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