desaichinmay22 wrote:
Hi Bunuel
The solution provided by you is the fastest. I tried to get the answer with a longer approach and couldn't get the right answer. Can please suggest where am I going wrong.
Area of floor = 3*2 = 6
Area of each block = 1*1 =1
No of blocks required = 6
We have 5 blocks of each white, black and red color.
Total color wise block arrangements that are possible to get from 6 blocks = 5+1, 4+1+1, 3+2+1 and 2+2+2
total ways to get blocks in (5+1) arrangement = 3C1*2C1*6!/5! = 3*2*6 = 36
total ways to get blocks in (4+1+1) arrangement = 3C1*6!/4! = 3*6*5 = 90
total ways to get blocks in (3+2+1) arrangement = 3C1*2C1*6!/2!3! = 3*2*6*5*4*3!/2*3! = 6*6*10 = 360
total ways to get blocks in (2+2+2) arrangement = 6!/2!2!2! = 6*5*4*3*2/2*2*2 = 30*3 = 90
Total = 36+90+360+90 = 576
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.
Hope it helps.
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