Ultrashort answer: 14C5 = 14C9.
This is a stars and bars problem. (Theorem 2 at the link)
You have nine bars
$$ ||||||||| $$
The spaces before, between, and after the bars correspond to the ten colors (where the $\cdots$ represent other colors)
$$ \text{orange}|\text{red}|\cdots|\cdots|\cdots|\cdots|\cdots|\cdots|\cdots|\text{black} $$
You distribute 5 stars among the bars, corresponding to a choice of numbers of jelly beans of the corresponding colors. For example
$$ \ast \ast |\ast||||||||\ast\ast $$
represents two orange, one red, and two black jelly beans.
So the revised question is "How many ways to arrange these $n$ stars and $k-1$ bars are there?" (There are $k = 10$ colors, but only $k-1 = 9$ bars by fenceposting, by our use of the spaces between the bars to represent colors.) Well, there are $n+k-1$ objects and choosing the positions of the stars completely specifies the locations of the bars (and vice versa), so there are $\binom{n+k-1}{n} = \binom{n+k-1}{k-1}$ arrangements of stars and bars. For $n = 5, k = 10$, this gives $\binom{14}{5} = \binom{14}{9}$ different bags of five jellybeans from 10 colors.
Many people choose to remember the form $\binom{n+k-1}{k-1}$ because the "$-1$" is common to both arguments in the binomial coefficient. There is another way to represent this number using multinomial coefficients, but that is not your question.
