Possible arrangements of marbles in bags? I've come across a question on a math test asking, "How many different ways can you put a dozen identical marbles into six bags so that each bag has atleat one marble in it?". I would imagine that this would be solved as if there were six marbles into six bags with no minimum per bag, but I don't know how to further solve this problem. The answer is 462. Any help would be appreciated.
 A: Hint: imagine the twelve marbles in a line. Place 5 vertical lines in the 11 interior gaps between pairs of adjacent marbles, no more than one line in any gap.
A: Hint: First put a marble in each bag. Then the question is reduced to how many ways are there of putting $6$ marbles into $6$ bags with no restrictions.
A: We can solve it by generating function. Since each bag can have a minimum of one marble and maximum of twelve, and there are six bags, 
the gf is:
\begin{align*}
  G(x) &= \left(x^{1}+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9}+x^{10}+x^{11}+x^{12}\right)^6
\end{align*}
and we need $[x^{12}]$, which can be extracted easily:
\begin{align*}
  G(x)&= \left(x^{1}+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9}+x^{10}+x^{11}+x^{12}\right)^6 \\
  &= x^6\left(\frac{1-x^{12}}{1-x}\right)^6 \\
  &= x^6\left(1-x^{12}\right)^6\sum_{k\ge 0}\binom{5+k}{k}x^k \\
  [x^{12}]G(x) &= \binom{5+6}{6} = 462
\end{align*}
A: As stated in the previous answers, once you put 6 marbles, 1 in each bag, the question is reduced to how many ways can you distribute 6 marbles into 6 bags, with no restrictions.
The way to solve this is imagining 5 dividers and 6 circles, where the circles represent the marbles, and the amount of marbles in between the dividers represents the amount of marbles you'd put in one of the bags, and because there are no restrictions, there can be no marbles in between 2 dividers. 
Now the problem is reduced to arranging 6 marbles and 5 dividers, which is ${11\choose6}$, which in fact is 462.
A: To distribute 6 marbles in 6 bags we just have to line up the 6 balls and then place 5 markers to indicate the possible distribution. For example using o for marble and | for a bag marker the sequence oooooo can be marked as oo||o|o|oo| indicating 2 in the first bag, none in the second bag, 1 in the third bag, 1 in the fourth bag, 2 in the fifth bag and none in the sixth bag. Then you see that we just need to choose the 5 marker positions from 11 possible positions. The number of ways is $\binom{11}{5}=462.$
