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13 May 2019, 01:20
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D
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Difficulty:
95%
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Question Stats:
36% (02:06) correct
64%
(02:25)
wrong
based on 139
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In the figure below, 3 of the 6 disks are to be painted blue, 2 are to be painted red, and 1 is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
A. 6
B. 8
C. 9
D. 12
E. 15
818c703b705b0707a63dee8af6f4b5c1d50be4a2.png [ 11.88 KiB | Viewed 6804 times ]
A. 6
B. 8
C. 9
D. 12
E. 15
Attachment:
818c703b705b0707a63dee8af6f4b5c1d50be4a2.png [ 11.88 KiB | Viewed 6804 times ]
09 Jun 2019, 08:17
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In the question, it is mentioned that 3 of the 6 disks are to be painted blue, 2 are to be painted red, and 1 is to be painted green.
Hence there are distinct groups of balls that will be after getting painted.
hence, we will solve in the following manner- 6!/(3!*2!*1!)=12
Since, on rotation or reflection the paintings will be considered the same, hence from any side, the value will be same which is 12.
Hence, answer is D.
Hence there are distinct groups of balls that will be after getting painted.
hence, we will solve in the following manner- 6!/(3!*2!*1!)=12
Since, on rotation or reflection the paintings will be considered the same, hence from any side, the value will be same which is 12.
Hence, answer is D.
ritwikkg
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07 Jul 2022, 22:16
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Aprajita760 Bunuel hakotaco
I did this:
6!
____________
6 x 3! x 2! x 1!
for circular arrangement
for similar objects
and this makes the answer 10
What is wrong in this approach?
I did this:
6!
____________
6 x 3! x 2! x 1!
for circular arrangement
for similar objects
and this makes the answer 10
What is wrong in this approach?
