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MOKSH
Joined: 01 Dec 2012
Last visit: 21 Sep 2019
Posts: 28
Given Kudos: 8
Concentration: Finance, Operations
Schools: Copenhagen '15 Iowa State '16 Jindal '15 (S) Haskayne(Calgary) AIM '16 SMU '15 Melbourne '17
GPA: 2.9
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Schools: Copenhagen '15 Iowa State '16 Jindal '15 (S) Haskayne(Calgary) AIM '16 SMU '15 Melbourne '17
Posts: 28
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:23) correct
36%
(02:26)
wrong
based on 335
sessions
History
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Not Attempted Yet
You are given an unlimited number of circles each of which having radii either 2 or 4. You must place the circles side-by-side so that they only touch at one point. If exactly two of the smaller circles are used, in how many different ways may the circles be placed next to one another so that the centers of the circles are all on the same straight line and so that the sum of the lengths of the diameters of the circles is 32?
A. four
B. six
C. eight
D. ten
E. twelve
A. four
B. six
C. eight
D. ten
E. twelve
11 Jan 2013, 04:18
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MOKSH
The the sum of the lengths of the diameters of two smaller circles is 2*4=8.
Hence, the sum sum of the lengths of the diameters of larger circles is 32-8=24, which means that there should be 24/8=3 large circles.
So, we have that there should be 2 small circles and 3 large circles.
The number of arrangement is 5!/(3!*2!)=10 (the number of arrangements of 5 objects in a row, where 3 of the objects are identical and the remaining 2 objects are identical as well).
Answer: D.
General Discussion
11 Jan 2013, 02:38
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MOKSH
We need to have following combination
1) 2 circles of radii 2
2) 3 circle of radii 4
Now...arrangements of N things in which p and q items are similar to each other.
5! / (3!*2!)
10
