Dropping letters in post boxes

Question

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?

Answer 1

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 $$(5,0,0),(0,5,0),(0,0,5)$$ $$(4,0,1),(4,1,0),(0,1,4),(0,4,1),(1,0,4),(1,4,0)$$ $$(3,1,1),(1,3,1),(1,1,3)$$ $$(2,3,0),(2,0,3),(3.0,2),(3,2,0),(0,2,3),(0,3,2)$$ $$(2,2,1),(2,1,2,(1,2,2)$$

Answer 2

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 $${4 \choose 2}=6$$

Answer 3

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