Permutation and combination: Distribution of toys among students 
In how many ways can $15$ toys be distributed among $3$ children if
  any child can take any number of toys ?
a) toys are distinct
b) toys are identical.

For the $b)$ Identical case:
Lets take $2$ bars diving $15 $toys into $3$ sections, so there are $16$ positions for every bar. So, since a student can get any number of toys (including $0$) The number of positions for the first bar is $16$ and that for the second bar is also $16$, 
Therefore, number of ways= $16*16=256$
Is this correct? And how to do the first part?
 A: Assuming giving the toys to different children is a different distribution:
For part $a$, consider the problem from the toys perspective. Each one has $3$ choices as to which child to be played with, and there are $15$ such choices to be made. Hence $3^{15}$ total options.
For $b$, as the other answer explained, the correct solution would be $\binom {17} 2$. Another way to get this: assume your dividers are different symbols - so order of their placement matters. Then there are $16$ (as you identified) ways to place the $1$st, but $17$ ways to place the second (as there are now $16$ symbols). In actuality, we don't distinguish between the symbols, so divide by $2! = 2$ - the result is $16 \cdot 17 / 2 = \binom {17} 2$
A: 
In how many ways can $15$ distinct toys be distributed to three children?

For each of the $15$ toys, we have three choices as to which child will receive the toy.  Therefore, there are $3^{15}$ ways to distribute the toys.

In how many ways can $15$ identical toys be distributed to three children?

Let $x_k$ be the number of toys received by the $k$th child, where $1 \leq k \leq 3$.  Then the number of ways the toys can be distributed is the number of solutions of the equation 
$$x_1 + x_2 + x_3 = 15$$
in the nonnegative integers.  A particular solution corresponds to the placement of two addition signs in a row of $15$ ones.  For instance, 
$$1 1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $x_1 = x_2 = x_3 = 5$, while 
$$1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 +$$
corresponds to the solution $x_1 = 10$, $x_2 = 5$, $x_3 = 0$.  Therefore, the number of ways of distributing $15$ identical toys to three children is the number of ways two addition signs can be inserted in a row of $15$ ones, which is 
$$\binom{15 + 2}{2} = \binom{17}{2} = 136$$
since we must choose which two of the seventeen positions (for $15$ ones and two addition signs) will be filled with addition signs.
