very stupid confusion in probablity or combination? i confuse very much on these when any question from probability is asked for example 
suppose you have three beggars let they be x,y,z and 2 biscuits and you have to distribute between them so total number of ways should be the numbers of solution of equation x+y+z=2 which is 6 and i know it i am certain of that but if we look at this way 

consider first biscuit it has three options and the other one it also has three options so it should be 3*3 that is 9 but it isn't i know it isn't but i am still very confused what it means mathematically and practically 

to explain what i am talking about consider these question i am not making these up they are in my textbooks

1 find the probability that a particular beggar gets exactly 1 biscuit 

2 number of ways of distributing biscuits between beggars

now on the first one the solution uses 9 as total number of ways but in second its 6 which is according to my intuition and which i feel should be used in first too ... what is going on here can someone explain me please examples of both cases with proper logic would be much appreciated
The number of non-negative integer solutions of the system:

$\begin{cases}x_1+x_2+\dots+x_r=n\\
0\leq x_1\\
\vdots\\
0\leq x_r\end{cases}$

is known to be $\binom{n+r-1}{r-1}$ and can be proven using a stars and bars argument.

Assuming the biscuits are considered identical and may not be broken apart, the number of ways of distributing the biscuits is then $\binom{2+3-1}{3-1}=\binom{4}{2}=6$ as you correctly found.

However: assuming each biscuit is given to one of the beggars uniformly at random, note that these six possibilities are not equiprobable.  It is twice as likely to arrive at the scenario $(1,1,0)$ than you are to arrive at the scenario $(2,0,0)$ (where the numbers correspond to the number of biscuits received by the respective beggar).

To answer the question of the probability, again assuming that the biscuits are distributed uniformly at random (this really should have been specified in the problem.  Perhaps in reality the shortest man is more likely to receive the biscuits since he looks more pathetic), imagine for a moment that the biscuits are in fact different colors.

You have nine different ways of distributing the biscuits now, and each of these are equiprobable.  The probability that the first beggar receives exactly one biscuit is then $\frac{1}{3}\cdot \frac{2}{3}+\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{9}$  (he either receives the red biscuit but not the blue one or he doesn't receive the red biscuit but does receive the blue one).

Remember that the scenarios $(\color{red}{1},\color{blue}{1},0)$ and $(\color{blue}{1},\color{red}{1},0)$ are considered different in this case, whereas in the other problem where the biscuits are considered identical.
To calculate probability: (no. of favorable outcomes)/(total no. of outcomes).

Probability that beggar gets exactly 1 biscuit is calculated as follows:

Total no. of favorable outcomes are : 6 (3 * 2 first biscuit can      be 
distributed in 3 ways while second biscuit can be distributed in 2 ways
since  none beggar can get more than one)

Total no. of outcomes : 9( 3 * 3 since first biscuit can be distributed in
3 ways and second biscuit can be distributed in 3 ways and each beggar can 
get more than one biscuit )

So probability : 6/9