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)

120 ways

Step-by-step 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.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

26?

step-by-step explanation: