permutation and combination confusion Can you point out where am I going wrong ?
I wanted to select 5 men from  7 men . This can be done very easily as 7C5 ways = 21ways, but I was confused as to why the following calculation didn't work out the same way ->
I first selected 3 men from 7 men as 7C3 ways = 35 ways
And then I selected 2 men from remaining 4 men as 4C2 ways = 6 ways ...so since I am selecting 3men AND 2men so 35x6= 210 
 A: You are overcounting in the following way. 
If the men were elements of $M = \{A,B,C,D,E,F,G\}$ then you might choose $A,B,C$ from $M$ and then $D,E$ from the remaining four or you might choose $A,B,D$ and $C,E$ from the remaining four but you get the same five men in both cases. 
For any given five element set there are $\binom{5}{3}=\binom{5}{2}=10$ ways to choose the first three elements and therefore you are counting each individual five element set $10$ times, as your calculations have already shown.
A: Let's try it with smaller numbers. I want to select $3$ men from $4$ men. This can be done easily in $_4\text{C}_3=4$ ways. (Equivalently, I can choose to omit one of the $4$ men in $_4\text{C}_1=4$ ways.) This is clearly the right answer.
Now let's try it another way. First I select $1$ man from $4$ men in $_4\text{C}_1=4$ ways. Then I select $1$ man from the remaining $3$ men in $_3\text{C}_1=3$ ways. Finally I select $1$ man from the remaining $2$ men in $_2\text{C}_1=2$ ways. So there are $4\times3\times2=24$ ways to select the $3$ men! Can you see why the second solution is wrong? By the way, how many ways are there to make an ordered selection (permutation) of $3$ men from the set of $4$ men?
A: Hi lets us concider a small real life example!
See....
let us concider that we have 5 members such as A B C D E ok.
Now you are willing to choose any 3 members from them to form a team.So then you will use the combination concept among them.right!fine...this will be simply 5C3 = ((120)/((2!)*(3!)))= 10.
But according to you "WHAT IF WE CHOOSE 2 MEMBERS FROM THESE 5 MEN AND THEN WHAT IF WE CHOOSE THE REMAINING 1 MAN FROM THE REMAINING 3 MEMBERS....CAN WE CHOOSE TOTAL 3 MEMBERS THIS WAY?AS WE CHOOSE 3 MEMBERS ABOVE IN PREVIOUS CASE CAN WE CHOOSE 3 MEMBERS FROM THESE 5 MEN BY APPLYING THE 2ND CONCEPT???according to you the answer gonna be (5C2)x(3C1)=30!!
*ok then let us now have anexperiment!!! :-) *
so what is the experiment!! well!!
      we have A B C D E as our 5 men ok.fine.then we are going to choose 3 members from them applying your concept and see what is going to be the result!! :-)
      first we  now choose 2 men from these 5 men!this can be done in 5C2 ways.that is 10 actually....ok?
now we can tabulate the pairs of the selected men....right!
 they may be:    {A-B , A-C , A-D , A-E , B-C , ... , D-E} ---- here is total 10 numbers of pairs we have got.fine!
     And now we will choose the remainin 1 member from the remaining 3 men.(cause our moto was to choose total 3 men from 5 members)
  now if we choose  that 1 guy from the remaining 3 men we will have only 3 choice!very simple!....but if we tabulate the choices now we will have: 

            {(C/D/E) , (B/D/E) , (B/C/E) , (B/C/D) , (A/D/E) , ... , (A/B/C)}---I think this line is not clear enough!ok let me make this clear to you..

here we write my 2 tabulated data in a readable manner:                       
pairs may get selected first: { A-B ,   A-C ,   A-D ,  A-E, ... ,  D-E}----(total 10 pairs)
choices for the last guy  :{ (C/D/E), (B/D/E),(B/C/E),(B/C/D)... (A/B/C)---(total 10 triplets of the alphabets for the corresponding pair selected in the first step!)
(please have a look into the parenthesis and the PAIRS and the respective choices in form of TRIPLETS)
I mean if (A-B) is selected then the choices are from the triplet of (C/D/E) -- (1)
         if (A-C) is selected then the choices are from the triplet of (B/D/E) -- (2)
         if (A-D) is selected then the choices are from the triplet of (B/C/E) -- (3)
         if (A-E) is selected then the choices are from the triplet of (B/C/D) -- (4)
                  ........and so on and atlast we have the combination            ...
         if (D-E) is selected then the choices are from the triplet of (A/B/C) -- (10)
so from the above experiment we can say,if the pair (A-B) is selected then C - D - E may be combined with A - B and we will get total (A-B-C) or (A-B-D) or (A-B-E) 
if it is A-C the pair then we will get total (A-C-B) or (A-C-D) or (A-C-E) like this we can proceed to fom our team of 3 members!!!
**But see here we have a case which is not admissible that is in these to pair of selection as (A-B) or (A-C) we have the team as (A-B-C) and (A-C-B)!!!!!!
if we apply your concept of counting as: (5C2)x(3C1)=30 we are counting these cases as two different teams or as two different selections which is not according to the concept of COMBINATION at all,the selections are basically same but we are counting them as two different team!!!did you get my point? :-)
here I am showing only one contradiction occurs in our experiment if we follow your concept of selection!!
Hope this discussion will help you to clear your doubt on the concept of COMBINATION or selection of objects from a given set!!best of luck... :-)
