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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.

 May 11, 2022
 #1
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+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, \(20+10+30+30=90 ways\)

 May 12, 2022
 #2
avatar+117491 
+1

mmm interesting use of LaTex.

 

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

 

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   wink

 May 12, 2022
 #4
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+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
avatar+117491 
+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
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+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
avatar+117491 
+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
 #8
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+2

Oooh!!! I get it now!!!

Thank you very much!!!!!! 

Guest May 12, 2022
 #9
avatar+117491 
+1

It's a pleasure :)

Melody  May 12, 2022
 #3
avatar+117491 
+1

I would like you to become a member.

We need people who are seriously trying to understand :)

If you get a good reputation your questions will get priority.

 May 12, 2022

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