How many different arrangements of letters are possible if three letters are chosen from the letters A through E and the letters E and A must be among the letters selected?
As A and E must be among 3 letters than the third letter must be out of B, C and D. 3C1=3 ways to choose which one it'll be. Now, 3 different letters can be arranged in 3!=6 ways, so final answer is 3*6=18.
Could we also solve this with: Total Combinations - Forbidden Combinations?
Total = 5*4*3 = 60
Forbidden (A is not part): 4*3*2 = 24
Forbidden (B is not part): 4*3*2 = 24
Forbidden (A and B are not part): 3*2*1 = 6
Total Forbidden Combinations: 54, Answer 6
I know its wrong but where is my mistake?
Help a KUDOS addict by clicking +1 Kudos and make me happy
PS Please send me PM if I do not respond to your question within 24 hours.