Let's take this hypothetical situation:

The deck has 6 cards:
Qc Kc Ac
Qs Ks As

Now, a flop can give: 3c, 2c 1s, 1c 2s, 3s.

Odds for 3c: (3/6)(3/6)(3/6)=(1/2)^3=1/8
Odds for 3s: 1/8 (same)

Now, what are the odds for 2c 1s (2 clubs and 1 spade)? According to your theory that the position doesn't matter, this would be:

(3/6)(3/6) [these are the clubs] x (3/6) [this is the spade]
=1/8

Now, 1c 2s has the same odds, so again 1/8.

So we add all the possibilities and we get 4/8=1/2. Where's the other 50%?

So yeah, position matters. In the 2c 1s example, there are three configurations, ie three places the spade can be at. So we have to multiply this by 3. Same for 1c 2s.

Thus we get: 1/8 for 3s and 2c; 3/8 for 2c 1s and 1c 2s

This adds up to 1.


Or to make a long story short.. you really do need to look at all the possible configurations of fitting the unsuited inside the suited cards. With 4 suited and 1 unsuited, there are 5. At the example of 3 suited and 2 unsuited, there are 10 such configurations.