9.8k views
2 votes
How many ways are there for Alice, Bob, Charlie, David and Eve to line up from left to right, such that Charlie is somewhere to the left of Eve (but not necessarily immediately to the left)

1 Answer

6 votes

Answer:

120 ways

Explanation:

Alice, Bob, David, Charlie, Eve

David, Alice, Bob, Charlie, Eve

Alice,David, Bob, Charlie, Eve

Charlie, Eve ,Alice,David, Bob,

Charlie, Eve ,Alice, Bob, David,

Charlie, Eve ,David, Alice, Bob,

Alice,Charlie, Eve , Bob, David,

Alice,Charlie, Eve , David, Bob,

Bob,Alice,Charlie, Eve , David, and so .

This is a permutation question as the order of placing Charlie to the left of Eve is important.

So the total number of people n= 5 and the possible order is 4 keeping Charlie left of Eve. Eve cannot have the last position to keep Charlie on the left.

Using the formula of nPr = n!/ (n-r)! we get

5! / (5-4)! = 120 ways in which Charlie can be placed to the left of Eve.

User Rager
by
7.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.