Question

http://screencast.com/t/K8nINdZO3go

@rational @Miracrown

Answer 1

well do you have notes from class on how to do this type of problem?

Answer 2

Yeah like take cases?

Answer 3

I'm not sure what you man by take cases, but I was thinking you might have notes on combinations or permutations?

Answer 4

Yes I have notes. By taking cases I mean all are different, 2 diff 2 same 3 same 1 diff.

Answer 5

This problem seems to be a little more involved than permutations and combinations then, is that correct? O.o

Answer 6

Nope. Simple combinations problem by taking cases.

Answer 7

I want to make sure that this problem is not more involved
I can help you if you just need to calculate a combination or permutation
so have you determined whether you will need to calculate a permutation or combination?

Answer 8

Here is the solution

http://screencast.com/t/JihpefJ9

Answer 9

What do u think now?

Answer 10

It doesn't open >.< !

Answer 11

o.O

Answer 12

http://screencast.com/t/JihpefJ9

Answer 13

IT OPENED!
FINALLY... :P

Answer 14

lol :P

Answer 15

It looks like they do not care about order (permutations). This is a combinations.
They are only interested in groups of letters.

Answer 16

what is it that you want to understand in that solution?

Answer 17

(i) Suppose all 4 are different.
Then what letters can the group contain (so that they are all different)
Maybe you should tell me which part of this explanation does not make sense to you.

Answer 18

I understand part 1

the rest?

Answer 19

OK O.O good that at least you've managed to understood part 1 :P
Are you also okay with the combinations of taking 4 things 4 at a time? 4C4 = 1

Answer 20

This is the number of ways we can select 4 items from 4 items.

Answer 21

Well I am not as smart as you @Miracrown or @AravindG :P

Answer 22

because the second part deals with 2 same, you don't even have to look at the o's. Just consider the number of ways in which you can select two letters other than the o's. You have three to select from : C,R and G. So you get 3C2.

Answer 23

Yes I know 4C4=1

Answer 24

lol @AravindG

Answer 25

Then if all 4 are not different, we could have 2 alike and 2 different.

Answer 26

The only letter that is repeated is O.

Answer 27

you know, posting the solution really kinda destroys the purpose. Now you wish to see how that was solved, instead of wondering how to solve it.

Answer 28

How many O's are there?
CORGOO

Answer 29

How we get 2c2?

Answer 30

@Miracrown I understood how we get 3 C2 How about 2C2 ?

Answer 31

good, that is from the other two letters that are not O's

Answer 32

CRG are there isnt it then y 2 not 3?

Answer 33

Wait. I see your question. I think that the 3 letters come from the different ones (not the O's) I think I was in error saying that above.

Answer 34

But then why 2C2 in the case of 0 o.O

Answer 35

If the O's are the same, we take 2 from 2.
3C2 refers to taking 2 letters from C, R and G.

Answer 36

If the O's are the same, we take 2 from 2.
Sorry didn't get that.

Answer 37

2C2 refers to taking 2 O's from 2....It is because they are the same letter.
It doesn't matter which pair of O's we take . They are indistinguishable.

Answer 38

2 0's 3 0's isnt it?

Answer 39

@OptimusPrime_ How about not looking at that solution pal? :D

Answer 40

@nipunmalhotra93 I am not looking at it.

Answer 41

lol I understand the question.
I don't know if I have a good answer for it. But my sense is that the 3C2 refers to the different letters C, R, G and the 2C2 refers to taking 2 from the O's (the same letter)

Answer 42

even though there are 3 of them.

Answer 43

o.O Then why not take it as 1 as all 3 are identical

Answer 44

When you are taking the case with 2 same, you don't have to look at the O's at all. You have three choices to select 2 letters from. That'll give you 3C2.

Answer 45

That's because the other two letters are already decided.

Answer 46

We can write out the sample space and see that the answer is correct.

Answer 47

\[\binom{3}{3} + \binom{3}{2} + \binom{3}{1} \]

Answer 48

following nipun's argument :
\(\binom{3}{3}\) is the number of ways u can choose 3 other letters (exclude O)

Answer 49

@Miracrown The answer is correct. But that solution doesn't seem to be.

Answer 50

O : \(\binom{3}{3}\)
OO : \(\binom{3}{2}\)
OOO : \(\binom{3}{1}\)

Answer 51

add them all to get the total combinations

Answer 52

We have 3 ways to select the 2 letters from C, R, G. The other two letters are O which has only one way to select 2 O's.

Answer 53

@rational is correct.

Answer 54

"That's because the other two letters are already decided. " lol that was the reply I was looking for :D

Answer 55

Okay looks good now.

Answer 56

wonder if there exists a general formula for problems like these...

Answer 57

like we have it for repeated permutations

Answer 58

Yeah.

Answer 59

stars and bars is not appropriate here i think...