The word "CROCODILE"

How many ways can we make three groups such that, each group consists of three letters, and the "Cs" are not together in the same group.

Guest May 11, 2022

#1**+1 **

Can someone please check this solution:

\(\text{C _ _ C _ _ _ _ _ }\)

Cases:

1: \(\text{C O _ C O _ _ _ _ } \implies 5C1*4C1*3C3=20\) ways.

2: \(\text{ C O O C _ _ _ _ _ }\implies 5C2*3C3=10\) ways.

3: \(\text{ C O _ C _ _ O _ _ } \implies 5C1*4C2*2C2=30\) ways.

4: \(\text{ C _ _ C _ _ O O _ } \implies 5C2*3C2*1C1=30\) ways.

Therefore, \(20+10+30+30=90 ways\)

Guest May 12, 2022

#2**+1 **

mmm interesting use of LaTex.

I will try to talk of your solution. You have made a very good attempt

1) CO _ CO_ _ _ _ You got 20 here. I think you need to divide that by two becasuse

for example COD COL *** is the same as COL COD ***

so I think this should be 10

2) COO C_ _ _ _ _ You got 10, I agree

3) CO_ C_ _ O_ _ We agree on 30

4) C_ _ C_ _ OO_ You got 30 I would halve that for the same reason as in 1. So 15 I think.

I think that covers all the cases so I get 10+10+30+15 = 65 ways.

I am not going to guarentee that it is correct though

Melody May 12, 2022

#4**+1 **

For 1):

But how would COD COL *** and COL COD *** occur?

To clarify: in the first space, we have 5 letters and we want to select 1 letter so 5C1 (This will give one of the following letters: R D I L E) Suppose it gave D. Then, in the second space, we have 4 letters now (R I L E) and we select 1 (for example, L) so 4C1, and the last 3 letters ( R I E ) in the last group.

So, how would C O L C O D *** appear ? ( I think It would appear if I said 5C1 and 5C1 again (5C1*5C1) but I removed a letter (5--->4))

By the way, 65 was also an answer I got randomly by other methods (but can't see the logic behind these, also the answer 105 and 210,420 etc.. all have some mistakes such as not accounting for the identical "Os" , and for 65 the above ^ reason is what I do not understand yet, I mean if it is possible that COD COL *** and COL COD *** would occur by case (1) then definitely, 65 is the correct answer! but I do not see how would these both occur at the same time by the same case.

anyway, this question is really silly :( ...

Thank you :)!!

Guest May 12, 2022

edited by
Guest
May 12, 2022

#5**+1 **

Maybe I do not understand your question but

COD COL RIE is the same as COL COD RIE You have counted it twice.

I think 65 is correct.

Melody
May 12, 2022

#6**+2 **

Yes it is the same but how would this occur (how would I have counted it twice If i used 5C1 then 4C1)

Guest May 12, 2022

#7**+1 **

Your 5C1 gives you 5 choices DLRIE

If you chose L then your 4C1 will give you the 4 coices DRIE say you then chose D

you now have COL COD RIE

alternatively you could have first chosen D and then chosen L

then you would have COD COL RIE

But these are the same, you have counted this outcome twice.

Melody
May 12, 2022