Permutation and CombinationHard
Question
In a conference 10 speakers are present. If S1 wants to speak before S2 & S2 wants to speak after
S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is :
S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is :
Options
A.10C3
B.10P8
C.10P3
D.10!/3
Solution
First we select 3 speaker out of 10 speaker and put in any way and rest are no restriction i.e. total number of ways
= 10C3 . 7.2! = 10!/3
= 10C3 . 7.2! = 10!/3
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