which one is correct? pls help me to verify!

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[Chong Lee Khim]

Dear teachers,

The question below is one of the questions from Trial exam paper (add maths paper 1).

   T  R  I  A  N  G  L  E  S
   

   Diagram 8 shows nine letters cards. Find the numbers of  possible arrangements of all the cards if
   
(a)   there is no restriction.      
 (b)   all the vowels must be separated
                             [4 marks]                         
                                                                                                                 
Answer from the marking scheme is as below:

(a)     9! = 362880
(b)
         __ C__C __ C __ C ___ C __ C ___ (create 7 space in between the consonants and insert 3 vowels)
            
      6! X ⁷P₂
         or                       
      = 720 x 210                         
      = 151200      

I tried other solution as below:

No. of arrangements = no. of arrangements without restriction – no. of arrangements with at least 2 vowels are placed together

No. of arrangements with at least 2 vowels are placed together: 

   VV  V  C  C  C  C  C  C

Merge 2 vowels to form one object:
   no. of ways = 8! X ³P₂

Thus, no. of arrangements = 9! - 8! X 3P2 = 120 960

Why the answer is not the same as the marking scheme’s one?  I was confused and felt hesitate to the working using the 7 space in between the consonants.  Can we accept the method like that?  I think the method using 7 space in between consonants means no. of ways to arrange 6 consonants X the no. of ways to arrange 7 objects (3 vowels and 4 empty cells), the no. of ways may include the arrangement of 6 consonants with empty cells and 2 or 1 vowel(s).
 
Please help me to clarify.  Thanks.

[Chong Lee Khim]

well, finally i found the mistakes!

when i tried to find the no. of arrangements with at least 2 vowels together, i accidently counted the no.of ways with 3 vowels together twice!  In the set of "at least 2 vowels", the set of "3 vowels together" is the subset.

Therefore, the proper working is
   9! - (8! x ³P₂ - 7! x 3!) = 151200  -- 7! x 3! is the no. of ways with 3 vowels together

 

[Justin Ngo]

yes.
this type of question when mentioned about separately, we must use the total minus the one they are side by side.

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