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6! ways (6*5*4*3*2*1 - since color once used can't be used for other sides).. IMO arijul007 wrote:In how many ways a cube can be colored using 6 colors if each color should be used atleast once?
A. 720
B. 120
C. 20
D. 30
E. 60
rijul007 wrote:In how many ways a cube can be colored using 6 colors if each color should be used atleast once?
A. 720
B. 120
C. 20
D. 30
E. 60
If I understand your solution, it doesn't look like you're including the constraint that every color has to be used at least once. Also, there would be more than one way to arrange the colors in most cases once you choose them.pemdas wrote:surprised with d thought it's e
with 6 colors applied individually it will make 6 ways
then 2 colors can be applied out of 6 colors, C(6,2) or 15 ways
additionally 3,4,5 and 6 out of 6 colors can be applied with the combination sets of 20,15,3 and 1 ways respectively.
in total 60 ways (6+15+20+15+3+1)
Why 6*6?smackmartine wrote:IMO D
Applying simple counting method, we know that 6 faces are available for 6 colors. So total # of ways = 6*6 = 36
different interpretations are possible with different colors on the cube's sidesGmatMathPro wrote:Imagine any given valid way to paint the cube with a different color on each of the six faces.
yes, we are counting unique ways as we perform simple permutation here of each color among the six given P(6,1). I am not sure if this is useful here as the colors put in different orders was not accepted by me. I voted for the sets of different colors (1,2,3,4,5,6) selected out of possible 6 colors. But even then with permuting each color out of 6 we get 720 and answer A.So, when you do 6*5*4*3*2*1=720, that is going to count each unique way of painting it 24 times.
Cannot agree with this more. As it was correctly noted before, there are 720 unique arrangements and when we discount these arrangements by 24 we are losing some valid sets here.Now, if you put the cube on a table, you could put any of the 6 faces facing up, and you could rotate it 90 degrees 4 times around a vertical axis that connects the top and bottom faces. That's 6*4=24 different arrangements for that one way of painting it. Hence, we have to divide by 24. 720/24=30.
Pete, I applied the same concept: if a work that can be done in m different ways and another work can be done is n different ways, total no. of ways both of them can be done in m*n ways.GmatMathPro wrote:Why 6*6?smackmartine wrote:IMO D
Applying simple counting method, we know that 6 faces are available for 6 colors. So total # of ways = 6*6 = 36
But exactly what two things are you saying can be done in 6 ways each? 6 ways to choose a side of the cube and then 6 ways to choose a color? If so, that would only get one side painted, and only 6 of these 36 paintings would be unique, due to rotational symmetry.smackmartine wrote:Pete, I applied the same concept: if a work that can be done in m different ways and another work can be done is n different ways, total no. of ways both of them can be done in m*n ways.GmatMathPro wrote:Why 6*6?smackmartine wrote:IMO D
Applying simple counting method, we know that 6 faces are available for 6 colors. So total # of ways = 6*6 = 36
eg. to prepare a meal there are following items - 3 kinds of breads , 3 kinds of toppings and 2 kinds of drinks..so there can be 3.3.2 kinds of meal. Is n't this right?
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