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19 Feb 2021, 03:45
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C
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Difficulty:
95%
(hard)
Question Stats:
42% (03:00) correct
58%
(02:41)
wrong
based on 60
sessions
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Not Attempted Yet
2 American men; 2 British men; 2 Chinese men and one each of Dutch, Egyptian, French and German persons are to be seated for a round table conference. If the number of ways in which no two people of the same nationality are together given by r*(6!), what is the value of r ?
A. 61
B. 122
C. 244
D. 250
E. 272
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 61
B. 122
C. 244
D. 250
E. 272
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
19 Feb 2021, 05:24
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SOLUTION:
Number of ways they can be seated
=9!=504(6!)
2 Americans, 2 Chinese, 2 British, 4 others
Let, A=>2 Americans together
B=>2 British together
C=>2 Chinese together
∴ Number of ways in which at least 2people of same nationality are together
=A+B+C−[A∩B+B∩C+C∩A]−A∩B∩C
Number of ways in which 2 Americans sit together across a round table
=8(2!) and the same is for 2 British and 2 Chinese
Thus total ways = 3(8!2!)
Number of ways in which 2 Americans and 2 Britishers sit together across a round table
= (7!2!2!) and the same is for 2 British and 2 Chinese and for 2 Americans and 2 Chinese and sitting together
Thus total double intersections = 3(7!2!2!)
Number of ways when 2 Americans,2 British and 2 Chinese sit together (triple intersection) across a round table = (6!2!2!2!)
Thus, A+B+C−[A∩B+B∩C+C∩A]−A∩B∩C
=3(8!2!)−3(7!2!2!)−(6!2!2!2!)
=6!(260)
Thus,Required number of ways = 6![504−260]=244(6!) OPTION (C)[
Hope this helps
Devmitra Sen(Math)
Number of ways they can be seated
=9!=504(6!)
2 Americans, 2 Chinese, 2 British, 4 others
Let, A=>2 Americans together
B=>2 British together
C=>2 Chinese together
∴ Number of ways in which at least 2people of same nationality are together
=A+B+C−[A∩B+B∩C+C∩A]−A∩B∩C
Number of ways in which 2 Americans sit together across a round table
=8(2!) and the same is for 2 British and 2 Chinese
Thus total ways = 3(8!2!)
Number of ways in which 2 Americans and 2 Britishers sit together across a round table
= (7!2!2!) and the same is for 2 British and 2 Chinese and for 2 Americans and 2 Chinese and sitting together
Thus total double intersections = 3(7!2!2!)
Number of ways when 2 Americans,2 British and 2 Chinese sit together (triple intersection) across a round table = (6!2!2!2!)
Thus, A+B+C−[A∩B+B∩C+C∩A]−A∩B∩C
=3(8!2!)−3(7!2!2!)−(6!2!2!2!)
=6!(260)
Thus,Required number of ways = 6![504−260]=244(6!) OPTION (C)[
Hope this helps
Devmitra Sen(Math)
24 Oct 2023, 12:41
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