There are 10 persons to address a meeting.In how many ways can they address the gathering if A speaks before B,B before C,and C before D?

Question

anonymous · Answer

HERE says the answer is 151200

anonymous · Answer

i wonder how you got that answer because i got the answer 7C4*6

anonymous · Answer

NO i didnt get the answer it was in a book and solution was written alongside it..I just wanted to know how to get it..

anonymous · Answer

Its simply 10! / 4!

anonymous · Answer

i dont know abt permutations and comninations

kinggeorge · Answer

I'm getting $$10C4\cdot6! ={10! \over 4!(10-4)!}\cdot6!={10!\over4!}$$you have 10C4 choices of where to put A, B, C, D, and 6! permutations of the other 6.

anonymous · Answer

Look at it this way.
10 people can speak in any manner had the condition not been placed.
So total ways = 10!
Now we have a designated order for the four people.
Hence thier arrangement needs to be removd. To do this we divide by their possible permutations = 4!
So it is 10!/4!

anonymous · Answer

i knw i mean i dont knw the permutation and combination not even oits basics!!!

anonymous · Answer

Ohh. Look up this site then : http://www.academicearth.org/lectures/permutations-and-combinations-1

anonymous · Answer

thnk  u