You are here,

Help, Search
New Batches at TathaGat Delhi!
Re: permutation
by Parvesh Angrula - Monday, 3 September 2012, 06:35 PM
  Answer would be 840 only. You can do in different way,
A can be on places 1 to 5, you can treat each case differently and count the total number of ways, it would come out to be 35.

but for each of those permutations of ABC, rest 4 lectures can be arranged in 4! ways,
so the answer is 35*4! = 840
Show parent | Reply