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
Create a free account to view solution
View Solution FreeMore Permutation and Combination Questions
In how many way can a game of tennis be played from 3 men and 4 women when each team contains one man and one woman-...If n identical dice are rolled, then no. of possible out comes are -...Consider all the permutations of the word CORRUPTION. Number of words in which U appears prior to both R is equal to -...How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?...There are 13 players of cricket out of which 4 are bowlers. In how many ways a team of eleven be selected from them so a...