Dropping letters in post boxes In how many different ways can 5 letters be dropped in 3 different post boxes if any number of letters can  be dropped in all of the post boxes?
 A: The number of the ways is exactly the number of the non-negative integer solutions of the equation 
$$x_1+x_2+x_3=5$$
Where $x_j$ denotes the number of letters dropped in $j-th$ post box. If empty post boxes are allowed, then the number of the solutions is $${7 \choose 2}=\frac{7\cdot 6}{2}=21$$
while the number of natural solutions is 
$${4 \choose 2}=6$$
A: In general number of droppings of $k$ letters in $m$ boxes is
$$\sum_{x_1+x_2+...+x_m=k,0\leq x_i\leq k}1=\binom{m+k-1}{k}$$
in our case $m=3,k=5$
$$\sum_{x_1+x_2+x_3=5,0\leq x_i\leq 5}1=\binom{5+3-1}{5}=21$$
Below is the list of all droppings
$$(5,0,0),(0,5,0),(0,0,5)$$
$$(4,0,1),(4,1,0),(0,1,4),(0,4,1),(1,0,4),(1,4,0)$$
$$(3,1,1),(1,3,1),(1,1,3)$$
$$(2,3,0),(2,0,3),(3.0,2),(3,2,0),(0,2,3),(0,3,2)$$
$$(2,2,1),(2,1,2,(1,2,2)$$
A: The first letter can be posted in any of the 3 post boxes. Therefore, it has 3 choices.
Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes.
Therefore, the total number of ways the 5 letters can be posted in 3 boxes is 3*3*3*3*3
= 3^5
