SOLUTION: Roman has a bag containing marbles of four different colours. How many ways can he place four marbles in a row so that no two marbles adjacent to each other are of the same colour?

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Question 206861: Roman has a bag containing marbles of four different colours. How many ways can he place four marbles in a row so that no two marbles adjacent to each other are of the same colour?
(A) 24 (B) 108 (C) 120 (D) 144 (E) 256
Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!
Roman has a bag containing marbles of four different colours. How many ways can he place four marbles in a row so that no two marbles adjacent to each other are of the same colour?
(A) 24 (B) 108 (C) 120 (D) 144 (E) 256

1.  Cases where all 4 are different colors:
  
    Choose a color for the 1st marble 4 ways.  That's 4 ways.
    Choose a color for the 2nd marble 3 ways.  That's 4*3 ways.
    Choose a color for the 3rd marble 2 ways.  That's 4*3*2 ways.
    Choose a color for the 4th marble 1 way.   That's 4*3*2*1 ways.

    These cases account for 24 ways.

2. Cases where the 1st and 3rd are the same color and the 2nd and 4th
   are the same color.

   Choose a color for the 1st and 3rd.  That's 4 ways.
   Choose a color for the 2nd and 4th.  That's 4*3 ways.

   These cases accounts for 12 ways.

3. Cases where the 1st and 3rd are the same color and the 2nd and 4th
   are different colors.    

   Choose the color for the 1st and 3rd.  That's 4 ways
   Choose the color for the 2nd,  That's 4*3 ways.
   Choose the color for the 4th.  That's 4*3*2.

   These cases accounts for 24 ways.

4. Cases where the 2nd and 4th are the same color and the 1st and 3rd
   are different colors.    

   Choose the color for the 2nd and 4th.  That's 4 ways
   Choose the color for the 1st,  That's 4*3 ways.
   Choose the color for the 3rd.  That's 4*3*2.

   These cases accounts for 24 ways.


5. Cases where the 1st and 4th are the same color and the 2nd and 3rd
   are different colors. 

   Choose the color for the 1st and 4th.  That's 4 ways
   Choose the color for the 2nd,  That's 4*3 ways.
   Choose the color for the 3rd.  That's 4*3*2.

   These cases accounts for 24 ways.

Adding:

Case 1:  24
Case 2:  12
Case 3:  24
Case 4:  24
Case 5:  24
-----------
Total:  108

So the correct choice is (B) 108

Edwin
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