Question: 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

Answer: 
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.
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?