SOLUTION: It is hard to understand where to use permutation and where to use combination. I need an answer w/ examples. Reference guide in pdf format can also do the trick. Thanks! :)

A Combination is just Choosing without positioning in any certain order.
A Permutation is Positioning in a certain order after choosing.

If we have 5 things, {A,B,C,D,E}, we can Choose 3 in these 10 ways:

{A,B,C}. {A,B,D}, {A,B,E}, {A,C,D}, {A,C,E}, 
{A,D,E}, {B,C,D}, {B,C,E}, {B,D,E}, {C,D,E}

IT's 10 ways because C(5,3} =  =  = 10

But each of those Combinations can be Positioned in 6 orders:

{A,B,C} can be Positioned in these 6 ways: ABC, ACB, BAC, BCA, CAB, CBA
{A,B,D} can be Positioned in these 6 ways: ABD, ADB, BAD, BDA, DAB, DBA
{A,B,E} can be Positioned in these 6 ways: ABE, ACE, BAE, BEA, EAB, EBA
{A,C,D} can be Positioned in these 6 ways: ACD, ADC, CAD, CDA, DAC, DCA
{A,C,E} can be Positioned in these 6 ways: ACE, AEC, CAE, CEA, EAC, ECA
{A,D,E} can be Positioned in these 6 ways: ADE, AED, DAE, DEA, EAD, EDA
{B,C,D} can be Positioned in these 6 ways: BCD, BDC, CBD, CDB, DBC, DCB
{B,C,E} can be Positioned in these 6 ways: BCE, BEC, CBE, CEB, EBC, ECB
{B,D,E} can be Positioned in these 6 ways: BDE, BED, DBE, DEB, EBD, EDB
{C,D,E} can be Positioned in these 6 ways: CDE, CED, DCE, DEC, ECD, EDC

So there are 6·10 or 60 Permutations of 5 things taken 3 ar a time.

P(5,3) = 5·4·3 = 60

P(n,r) = C(n,r)·r!

"COMBINATION" starts with a C and "CHOOSE" starts with a C.

"PERMUTATION" starts with a P and "POSITION" starts with a P.

Actually in both cases you choose, but with Permutation you Position
each combination in all the possible orders.

Edwin