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18 Feb 2021, 03:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
33% (02:11) correct
67%
(02:25)
wrong
based on 83
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There are 12 guests at a dinner party. Supposing that the master and mistress of the house have fixed seats opposite one another, and that there are two specified guests who must always, be placed next to one another; the number of ways in which the company can be placed is :
(A) 20*10!
(B) 22*10!
(C) 44*10!
(D) 46*10!
(E) 48*10!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 20*10!
(B) 22*10!
(C) 44*10!
(D) 46*10!
(E) 48*10!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Neha2404
Joined: 07 Mar 2020
Last visit: 27 Oct 2025
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Concentration: Finance, Accounting
Schools: IIMA PGPX'22 ISB'22 IIMB EPGP'22 Kellogg '22
18 Feb 2021, 04:34
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IMO-A
Master and mistress can have seats in - 2 ways
2 guests who seat next to each other can have seat in -2*2*5 ways
Remaining 10 guest can have seats - 10!
Total no of ways is
2*2*5*10!
=20*10!
Posted from my mobile device
Master and mistress can have seats in - 2 ways
2 guests who seat next to each other can have seat in -2*2*5 ways
Remaining 10 guest can have seats - 10!
Total no of ways is
2*2*5*10!
=20*10!
Posted from my mobile device
18 Feb 2021, 08:26
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Along with the 2 hosts we have 12 guests. So there are 14 people.
Now if the master sits first, there is only 1 way that he can choose a seating arrangement as all seating arrangements are the same for him.
Similarly, once he is seated, there is only way the lady can sit, which is opposite to him.
Let us now take the 2 people sitting together. There will be 6 chairs on either side of the master and his wife. Let us say these are a,b,c,d,e and f on 1 side. Then the possible ways on one side is ab, bc, cd, de and ef which is 5 ways.
Similarly on the other side we will have 5 ways. Total ways is 10.
these 2 people can interchange (XY and YX) in 2! = 2 ways.
Total number of ways to seat these 2 people = 10 * 2 = 20 ways.
The remaining 10 guest can sit in 10! ways.
The total number of ways = 20 * 10!
Option A
Arun Kumar
Now if the master sits first, there is only 1 way that he can choose a seating arrangement as all seating arrangements are the same for him.
Similarly, once he is seated, there is only way the lady can sit, which is opposite to him.
Let us now take the 2 people sitting together. There will be 6 chairs on either side of the master and his wife. Let us say these are a,b,c,d,e and f on 1 side. Then the possible ways on one side is ab, bc, cd, de and ef which is 5 ways.
Similarly on the other side we will have 5 ways. Total ways is 10.
these 2 people can interchange (XY and YX) in 2! = 2 ways.
Total number of ways to seat these 2 people = 10 * 2 = 20 ways.
The remaining 10 guest can sit in 10! ways.
The total number of ways = 20 * 10!
Option A
Arun Kumar
