I got this question wrong. Need help to solve the problem. Guys, can you pls help me by giving an approach to attack this type of problem?

Wish to hear from you guys ...

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 ?

Case 2: Joey is the second person in the line. thus frankie can occupy 4 positions from X3 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 third person in the line. thus frankie can occupy 3 positions from X4 to X6.

this problem has a lot of mileage on it already in this forum.

just type "mobsters" into the search box, and you'll find the following:
http://www.manhattangmat.com/forums/six ... t1357.html
http://www.manhattangmat.com/forums/post2952.html

check those out. if you still have questions, then post them - preferably in one of those threads, as it's nice to have everything in one place.