12 people, including the MD, the CFO, and 10 executives are to be seat

Question

12 people, including the MD, the CFO, and 10 executives are to be seated around a circular table. Find the probability that there are at least 3 executives sitting between the MD and the CFO

A. 1/2
B. 3/5
C. 1/3
D. 5/11
E. 2/11

preetamsaha · Accepted Answer

let, the MD is positioned at point 1, 
then the CFO can be positioned at any of the points 5, 6, 7, 8, 9 (  CFO cannot be positioned at 2,3,4,10,11,12 otherwise the number of executives between MD and CFO will be less than 3 on one side  )
 Number of favourable positions = 5.
 Total number of available positions = 11 
Thus the required probability = 5/11.

correct answer is D

ArunSharma12 · Answer

Total number of arrangements of 12 people around a circular table =  (12-1)! 
when 3 employees sit between MD and CEO, arrangement looks like  MD-X-X-X-CEO 
first we will have to select these 3 employees out of 10 employees available =  10_C_3  
after having selected the 3 employees, there are 7 employees and the group  MD-X-X-X-CEO  left to arrange. consider this group as one entity.
number of ways of arranging 7 employees and this group around a circular table =  10_C_3*(7+1-1)!*3!*2 ; 
3! corresponds to number of ways of arranging three employees.
2 corresponds to two different arrangements by swapping the places for MD and CEO. 
Probability =  \frac{10_C_3*(7+1-1)!*3!*2}{11!}  =  \frac{10!*(7)!*3!*2}{7!*3!*11*10!}  =  \frac{2}{11}

GMATWhizTeam · Answer

Solution:  
We will solve this problem by eliminating the unfavorable case.
 https://www.dropbox.com/s/aj1l7xphqfvcbue/39.JPG?dl=1 
 o Total number of seating arrangements = 11!  
 
 o There are 3 unfavorable cases
o When no executive is seated between MD and CFO.
  Number of ways =  10!*2    
o When only one executive is seated between MD and CFO.
  Number of ways =  10C1 *9!*2 = 10!*2    
o When there are only two executives are seated between MD and CFO.
  Number of ways = 10C2*8!*2*2 = 10!*2    
o Total number of unfavorable case = 10!*2+ 10!*2 + 10!*2 = 6*10! 
o Probability that there are at most two executives between MD and CFO =  \frac{6*10!}{11!} = \frac{6}{11} 
o Probability that at least 3 executives are seating between MD and CFO =  1 – \frac{6}{11} = \frac{5}{11} . 
Hence, the correct answer is  Option D.

u2409 · Answer

For each of 12 positions of CFO there are 5 possible positions for MD, which makes them at least 3 position apart both ways.
If you imagine a clock, if CFO is positioned on 12 then MD can be anywhere between 4 and 8. Which are 5 distinct areas for MD to be. So total such arrangements of positions are 12*5.
And total distinct arrangements of positions of MD and CFO is 12*11.
So we divide 12*5 by 12*11 and get 5/11

12 people, including the MD, the CFO, and 10 executives are to be seat

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