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03 Mar 2021, 09:15
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D
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Difficulty:
75%
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Question Stats:
52% (01:58) correct
48%
(02:20)
wrong
based on 107
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In how many ways the numbers 1, 2, 3, 4, 5, and 6 can be arranged around a circle so that at least two odd numbers are together? (2 seating arrangements are considered different only when the positions of the people are different relative to each other.)
A. 12
B. 24
C. 36
D. 108
E. 120
A. 12
B. 24
C. 36
D. 108
E. 120
03 Mar 2021, 19:05
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The odd numbers are 1, 3 and 5.
At least 2 odd numbers together means exactly 2 odd numbers together + all 3 odd numbers together
We can have 1, 3 or 1,5 or 3,5 together. For each of these, let us consider the 2 odd numbers as 1 unit.
Arrange 5 units in a circle which can be done in 4! = 24 ways. The odd numbers can arrange internally in 2! = 2 ways.
Therefore arrangements with 2 odd numbers together = 3 * 24 * 2 = 144
What we must remember is that these 144 arrangements also contain those arrangements where the 3rd odd number can be on either side of these 2 odd numbers. Therefore for each of these 3 sets, we need to subtract the possibilities where the odd numbers come together.
To have all 3 odd numbers together. Take all of them as 1 unit. Then we need to arrange 4 units in a circle which can be done in 3! = 6 ways. These numbers can rearrange themselves internally in 3! = ways.
Arrangements having all 3 odd numbers together = 6 * 6 = 36
Total number of arrangements = 144 - 3*36 + 36 = 108
Option D
Arun Kumar
At least 2 odd numbers together means exactly 2 odd numbers together + all 3 odd numbers together
We can have 1, 3 or 1,5 or 3,5 together. For each of these, let us consider the 2 odd numbers as 1 unit.
Arrange 5 units in a circle which can be done in 4! = 24 ways. The odd numbers can arrange internally in 2! = 2 ways.
Therefore arrangements with 2 odd numbers together = 3 * 24 * 2 = 144
What we must remember is that these 144 arrangements also contain those arrangements where the 3rd odd number can be on either side of these 2 odd numbers. Therefore for each of these 3 sets, we need to subtract the possibilities where the odd numbers come together.
To have all 3 odd numbers together. Take all of them as 1 unit. Then we need to arrange 4 units in a circle which can be done in 3! = 6 ways. These numbers can rearrange themselves internally in 3! = ways.
Arrangements having all 3 odd numbers together = 6 * 6 = 36
Total number of arrangements = 144 - 3*36 + 36 = 108
Option D
Arun Kumar
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03 Mar 2021, 19:36
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CrackVerbalGMAT
I have a small doubt. why have you subtracted 3*36 and not simply 36 from the 144?
