Seven speakers A1, A2, A3,...., A7 were scheduled to speak at a function. In how many ways their speech can be arranged, such that:
A1 speaks before A3 , and A3 speaks before A5 .

a. 840
b. 860
c. 420
d. 410

All the seven speakers can be selected in 7 ways..=7!
But A1 should speak before A3 and A3 before A5 that=A1A2A3
=3!..hence it must ve excluded.
=7!/3!
=840
10 years agoHelpfull: Yes(9) No(8)
3 people can be selected from 7 in 7c3 ways
arranging those 3 people is only in one way.
remaining 4 can be arranged in 4! ways.
so 7c3*1*4!=840
short cut: 7!/3!
10 years agoHelpfull: Yes(9) No(0)
take the position of a1 a2 a3 by 7c3 ways and then remaining 4 arranged by 4! ways so the ans will be 7c3*4!
10 years agoHelpfull: Yes(4) No(0)
total number of arrangements is 840
10 years agoHelpfull: Yes(1) No(0)

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