Q A circle is divided into 4 parts.five colors are avaliable.In how many ways this cirlce can be colored if no two adjacent part should should have same color and repetition of colors is allowed.
a)625
b)120
c)240
d)320
e)none
ans is 320
counting probs help plz
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- kmittal82
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The circle is immaterial in this one, instead think of this question as how can you fill 4 boxes with 5 colours, with no 2 adjacent boxes with the same color. Here's a (bad) representation of the boxes
|5 |4 |4 |4 |
Now, first box can be filled in 5 ways, next one in 4 ways (since same colors cant be adjacent), next one again in 4 and same for the last one.
So, total number of ways = 5 x 4 x 4 x 4 = 320
|5 |4 |4 |4 |
Now, first box can be filled in 5 ways, next one in 4 ways (since same colors cant be adjacent), next one again in 4 and same for the last one.
So, total number of ways = 5 x 4 x 4 x 4 = 320
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dont u think it should be 5x4x4x3. at last quart we have 3 options because it has two adjacent quarts and colored used here can not be used
- kmittal82
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Thats a good question, and made me think as well. But consider this, lets say we have 5 colours, ABCDE.quantskillsgmat wrote:dont u think it should be 5x4x4x3. at last quart we have 3 options because it has two adjacent quarts and colored used here can not be used
Now, imagine filling in the circle in a clockwise fashion. First spot obviously can be 5 ways, next one 4, the one after another 4. As for the final piece, we could still have 4 options, since repetition is allowed.
E.g, if you fill in A,B,A, then you still have 4 options to fill the final pie, since the 2 colours next to the final pie will be the same.