A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted?(A) 104
(B) 213
(C) 577
(D) 705
(E) 726
There are 5 white, 5 black, and 5 red blocks available to fill 2*3=6 slots. Following 6 cases are possible for different pattern arrangements:
5-1: 5 blocks of the same color and 1 block of different color: \(C^1_3*C^1_2*\frac{6!}{5!}=36\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 5 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 1 blocks, and \(\frac{6!}{5!}\) is ways of different arrangements of 6 blocks, XXXXXY, out of which 5 are identical);
4-2: 4 blocks of the same color and 2 block of different color: \(C^1_3*C^1_2*\frac{6!}{4!*2!}=90\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 4 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 2 blocks, and \(\frac{6!}{4!*2!}\) is ways of different arrangements of 6 blocks, XXXXYY, out of which 4 X's and 2 Y's are identical);
The same way for other patterns:
4-1-1: \(C^1_3*\frac{6!}{4!}=90\);
3-3: \(C^2_3*\frac{6!}{3!*3!}=60\);
3-2-1: \(C^1_3*C^1_2*\frac{6!}{3!*2!}=360\);
2-2-2: \(\frac{6!}{2!*2*2!}=90\);
Total: 36+90+90+60+360+90=726.
Answer: E.
Shorter approach:Imagine the case in which we have not 5 blocks of each color but 6, then each slot from 2*3=6 would have 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 6 slots would be 3*3*3*3*3*3=3^6;
Now, what is the difference between this hypothetical case and the one in the question? As we allowed 6 blocks of each color instead of 5, then we would get 3 patterns which are impossible when we have 5 blocks of each color: all white, all red and all black. Thus we should subtract these 3 cases: 3^6-3=726.
Answer: E.
kt00381n wrote:
Why 15!/6!9! wont work? Im confused.
I you look at the first approach you'll see that \(C^6_{15}\) doesn't give all possible patterns possible. \(C^6_{15}\) is # of different groups of 6 possible out of 15 distinct objects, which clearly is not the case here.
diddygmat wrote:
This does not make sense. Are we not counting repeating patterns here? The question syas different pattern...
for example: The pattern BBBRRR is being counted at least a 2 times here. I s my understanding right?
Thanks!
Both approaches above count different patterns. Consider the following case: there are 2 slots to fill and 4 white, 4 black, and 4 red blocks available. How many different arrangements are possible:
WW
WB
BW
WR
RW
BB
BR
RB
RR
Total of 9 different arrangements. The same as if we consider approach #2 from above: each slot from 2 has 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 2 slots would be 3*3=3^2=9.
Hope it helps.