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A pentagonal prism is to be painted. No two adjacent sides [#permalink]
 04 Aug 2003, 20:18
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A pentagonal prism is to be painted. No two adjacent sides are painted, in how many ways can the prism be painted?
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4 Ways.
Just an educated guess. I need to know the mathematical way of solving the above problem.
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mistdew wrote: A pentagonal prism is to be painted. No two adjacent sides are painted, in how many ways can the prism be painted?
The first face to be painted can be chosen in 5 ways.
The second face can be chosen in only two ways since no adjacent sides are painted.
Therefore, total no of ways = 5*2=10
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prashant wrote: mistdew wrote: A pentagonal prism is to be painted. No two adjacent sides are painted, in how many ways can the prism be painted? The first face to be painted can be chosen in 5 ways. The second face can be chosen in only two ways since no adjacent sides are painted. Therefore, total no of ways = 5*2=10 A pentagonal prism has 7 faces. The top, the bottom, and five vertical faces. If we choose either the top or the bottom, we have only the opposite face to paint because other five faces are adjacent. Thus, 2 ways.
If we take each of vertical faces, we have two faces to paint. Thus, 5*2=10
Finally, 2+10=12
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stolyar wrote: prashant wrote: mistdew wrote: A pentagonal prism is to be painted. No two adjacent sides are painted, in how many ways can the prism be painted? The first face to be painted can be chosen in 5 ways. The second face can be chosen in only two ways since no adjacent sides are painted. Therefore, total no of ways = 5*2=10 A pentagonal prism has 7 faces. The top, the bottom, and five vertical faces. If we choose either the top or the bottom, we have only the opposite face to paint because other five faces are adjacent. Thus, 2 ways. If we take each of vertical faces, we have two faces to paint. Thus, 5*2=10 Finally, 2+10=12 If the problem does not state that the colors are to be different, you may be making the mistake of double counting. Let's say the side faces are numbers 1 through 5. Pick 1, you can combine with 3 and 4. Pick 3 you can combine with 5 and 1. You have just double counted 1 and 3 unless the problem specified that they are to be painted different colors. Same with the top and bottom.
Be careful.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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stolyar wrote: prashant wrote: mistdew wrote: A pentagonal prism is to be painted. No two adjacent sides are painted, in how many ways can the prism be painted? The first face to be painted can be chosen in 5 ways. The second face can be chosen in only two ways since no adjacent sides are painted. Therefore, total no of ways = 5*2=10 A pentagonal prism has 7 faces. The top, the bottom, and five vertical faces. If we choose either the top or the bottom, we have only the opposite face to paint because other five faces are adjacent. Thus, 2 ways. If we take each of vertical faces, we have two faces to paint. Thus, 5*2=10 Finally, 2+10=12 Stolyar,
I had thought of the top and the bottom face too. But I assumed that "adjacent" would mean any two faces that share an edge. If that's the case, then the top and botttom would always be adjacent to all the vertical faces, and consequently we cannot paint them. This was the basis for my answer. What do you think?
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why can't we paint the top and the bottom? From the verbiage of the question stem we can assume that we have the only paint and that we may paint any face.
Akamai is correct, we can paint the top and the bottom and so have the only combination.
As for side faces, a rotation factor should be eliminated. Fix the front face (1) -- we have two faces to paint (3) and (4) (counting clockwise and looking from the top). 2 combinations. For (2) we have (4) and (5). For (5) we have (2) and (3). Thus, overall, 6 combinations.
total 6+1=7
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If only one face is painted, then there are 7 ways of painting the prism.Two faces can be painted in the foll ways
Base & top - 1 way
Two vertical faces - not adjacent - can be painted in 5 ways
thus, total no of ways 7+1+5=13..
Right?
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