by Saurav Tue Oct 21, 2008 3:04 pm
I approach questions like these as follows (counting method)
Draw (!!!) the situation
X1------X2------X3------X4------X5------X6
where X is the position of the mobsters in the queue. Say the queue starts from left.
Condition is that Frankie (F) behind Joey (J) but not necessarily the next one behind him
Case 1: Joey is the first person in the line. thus frankie can take any position from X2 to X6.
J------X2------X3------X4------X5------X6
Joey can occupy X1 position in 1 way, Frankie can occupy 5 positions (X2 to X6) and the remaining 4 mobsters can occupy remaining 4 places in 4! ways.
So total number of ways = 1 x 5 x 4!
Case 2: Joey is the second person in the line. thus frankie can take any position from X2 to X6.
X1------J------X3------X4------X5------X6
Joey can occupy X2 position in 1 way, Frankie can occupy 4 positions (X3 to X6) and the remaining 4 mobsters can occupy remaining 4 places in 4! ways.
So total number of ways = 1 x 4 x 4!
Case 3: Joey is the first person in the line. thus frankie can take any position from X2 to X6.
X1------X2------J------X4------X5------X6
Joey can occupy X3 position in 1 way, Frankie can occupy 3 positions (X4 to X6) and the remaining 4 mobsters can occupy remaining 4 places in 4! ways.
So total number of ways = 1 x 3 x 4!
Case 4: X1------X2------X3------J------X5------X6
Joey can occupy X4 position in 1 way, Frankie can occupy 2 positions (X5 or X6) and the remaining 4 mobsters can occupy remaining 4 places in 4! ways.
So total number of ways = 1 x 2 x 4!
Case 5: X1------X2------X3------X4------J------X6
Joey can occupy X5 position in 1 way, Frankie can occupy 1 position (X6) and the remaining 4 mobsters can occupy remaining 4 places in 4! ways.
So total number of ways = 1 x 1 x 4!
Total 5 cases hence add all the cases
= 1 x 5 x 4! + 1 x 4 x 4! + 1 x 3 x 4! + 1 x 2 x 4! + 1 x 1 x 4!
= 1 x ( 5 +4 +3 +2 +1 ) x 4!
= 1 x 15 x 4! (=24)
= 360
Is 360 the answer ?
