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Numerical Ability
Permutation and Combination
15 people sit round a circular table what are the odds against two particular people sitting together.
(a)7:1
(b)6:1
(c)5:1
(d)1:5
Read Solution (Total 9)
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- ans is 6:1
odds agianst an event=
(possibilities of event not happenning)/(possibilities of event happening)
=possibilities that 2 people are not togeather/possi that they are togeather
= (14!-2*13!)/(2*13!)
= 6:1 - 11 years agoHelpfull: Yes(70) No(1)
- @SUJITHA MADDULA
in circular arrangement total no.of possiblities are (n-1)!
her two particular people should sit together
so we consider them as single person so total no.of people are 14
and no.of arrangements possible are (14-1)! = 13!
now the 2 people who are should be together can be arranged in 2 ways
so total no.of possiblilities that they are together are 2*13! - 11 years agoHelpfull: Yes(12) No(1)
- if n people are in circular fashion,the number of arrangements possible are
(n-1)! but not n!.So,the possi that the two are togeather is (14-1)!*2! but not 14!*2!@LAXMI - 11 years agoHelpfull: Yes(6) No(2)
- plz clear tis doubt @BAILORE...
2 people are sitting together then possi for that they are together is 14!*2! then how 13!*2!....
- 11 years agoHelpfull: Yes(5) No(4)
- odds against an event=no of unfavorable choices/favourable choices
((14!-2*13!)/(2*13!))=6/1=6:1 - 11 years agoHelpfull: Yes(3) No(1)
- can u explain what is the 2*13! @bailore pradyumna acharya
- 11 years agoHelpfull: Yes(2) No(3)
- the all independent people can sit themselves in 14! ways in circular arrangement
when 2 people sit together take as 1 unit the remaining 13+1 unit is arranged in 13 ways and 2 people among themselves 2! ways ratio of 1&2 we get 7:1 - 11 years agoHelpfull: Yes(1) No(1)
- ok..i understand the logic but can anybody explain me how (N-1)!number of ways are there for N people sitting in circular
- 11 years agoHelpfull: Yes(0) No(0)
- 7:1 because all people sit around a round table if any two particular people selected as sitting to gether remaining odds are only 7
- 8 years agoHelpfull: Yes(0) No(0)
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