1why nC2??
n! = no. of ways n people sit
(n-1) = no. of ways they sit 2gether
Suppose n (≥3) persons are sitting in a row. Two of them are selected at random. The probability that they're not together is
a) 1-(2/n)
b) 2/(n-1)
c) 1-(1/n)
d) 2/(n+1)
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13 Answers
49let us first of all select two people...that is in nC2 ways...then we find out the number of ways in which they can sit!! it is (n!)(nC2) ways!!
then next!! we select two people and make them sit such that they sit always together...it can be done in (n-1)! (nC2)
so probability the twopeople sit together is (n-1)! (nC2)(n!)(nC2)=1n
thus probability they are not sitting together=1-1n
11@sb
the 2 selected people can also rearrange amongst themselves.
so 4 sitting together P(A)=2/n
so ans. should be a)
49let us first of all select two people...that is in nC2 ways...then we find out the number of ways in which they can sit!! it is (n!)(nC2) ways!!
then next!! we select two people and make them sit such that they sit always together...it can be done in (n-1)! (nC2)X2!
so probability the twopeople sit together is (n-1)! (nC2)X2!(n!)(nC2)=2n
thus probability they are not sitting together=1-2n
49@SHANKAR
nC2 because we are selecting two people outta the n people ...[1][1][1]
1nC2 not necessary . Y do v select 2 people
n! = no. of ways n people sit
n-1! = no. of ways n people sit taken 2 (who r together) at a time
21yes nc2 is not necessary here as total ways in which n numbers can be arranged is n!......then consider two numbers as one object........so total no. of ways in which both nos are together is (n-1)!*2!
hence probability of both sitting together is 2/n and not sitting together is 1-2/n.
49yea..i think you may be correct!! this will be cnfusing as both gives same result but what about this part??
QUOTE:
Shankar said: ""...row. Two of them are selected at random. The pr...""