How many differnent ways to pick n marbles of five colors at random?

Question

Suppose you have bag of $p$ marbles with $a$ marbles of a color $A$, $b$ of color $B$, and so on to $e$ marbles of $E$ i.e. $p=a+b+c+d+e$. Consequentially for any singular type of marble e.g. $A$, $0\le a \le p$ so long as $p=a+b+c+d+e$.

While there are $\binom{p}{n}$ ways to pick $n$ marbles out of $p$ marbles, I am more interested in figuring out, if given $a$, $b$, $c$, $d$, and $e$ how many unique ways there are to do this. For example, suppose $n=2$, $p=10 = a =10, b=c=d=e=0$. There are $\binom{p}{n}$ ways for me to get two marbles, but no matter which two I choose I will have two marbles of type $A$, so there is really only one result.

I guess then my question is more so inverted. Given that I am looking to choose $n$ how many ways can I construct $n$ from $a, b, c, d$, and $e$.

My first thought was the multinomial coefficient

$$ \frac{n!}{a!b!c!d!e!} $$

but that clearly is incorrect. What am I forgetting? Looking at this question it seems that I have to do this for the variable number of ways this could occur:

$$ 5(\binom{n}{n} + \binom{n}{n-1}*4\binom{n}{n-n+1} + \binom{n}{n-2}*8\binom{n}{n-n+1}*4\binom{n}{n-n+2} + \dots) $$ but that only works assuming that there are $n$ of each type of marble....

Answer 1

You need to be crystal clear about the multinomial coefficients.

For $10$ marbles, if you want $4$ of a, $3$ of b and one each of $c,d,e,$

the multinomial coefficient $\dfrac{10!}{4!3!1!1!1!}$ indeed gives the answer you desire,

but if you instead specify $3$ of one color, $2$ of another, and $1$ each of the remaining colors, you need to multiply the above by another multinomial coefficient, $\dfrac{5!}{1!1!3!}$ to take into account that the pattern could be generated with different colors in many ways.

The answer could, of course, be abbreviated as $\dfrac{10!}{4!3!}\cdot\dfrac{5!}{3!}$

but I recommend the full form including $1!$ and (if it occurs), $0!$ to give a sort of double check that the numerator and denominator tally.

---

Addendum

Supposing you want at least one of each color, you will first have to form all patterns, eg $6,1,1,1,1\quad5,2,1,1,1\quad4,3,1,1,1\;$(the pattern we discussed), and continue down to $2,2,2,2,2$, compute separately for each case and add up

But there is a shortcut using inclusion-exclusion, viz

$5^{10} - \binom514^{10} + \binom523^{10} - \binom532^{10} + \binom541^{10}$

An even shorter way (if you know about Stirling numbers of the second kind, and have tables of such numbers) is to compute $S2(10,5)*5!$

Answer 2

In terms of generating functions the number you are looking for is: $$ [x^n]\prod_{i=1}^k \sum_{j=0}^{N_i}{x^j}, $$ where $k$ is the number of colors, $N_i$ is the number of marbles of the $i$-th color, and $[x^n]$ denotes the coefficient at $x^n$ in the power expansion of the function.

Answer 3

Adding some explanation to the answer of @user.

Let us embed the question into a Ordinary Generating Function (OGF). See this related question for EGF in case order matters to you: 4 digit combinations of 12333210, 3 letter combinations of MISSISSIPPI

Simpler Problem: What you need is to find the coefficient of $x^n$ in

$$(1+x+x^2+\dots+x^a)\times(1+x+x^2+\dots+x^b)\times\dots\times(1+x+x^2+\dots+x^e)$$

Explanation: The number of ways to pick n marbles of the 5 types, is the same as the number of ways to pick n $x's$ from each of the 5 polynomial sub-expressions. This embedding ensures,

• You can't pick more than $a$ number of x's of type A, and similarly for the other types.
• You pick a total of n $x$'s across the generating polynomials (as you want coefficient of $x^n$).
• You cannot pick a negative number of any type.
• The expression is unique and therefore well-defined.
• Each way of picking $x^n$ from the 5 sub-polynomials adds 1 to the coefficient of $x^n$, and the total number of ways of picking $x^n$ is therefore the coefficient of $x^n$ in the full expression.

Thus the embedding is valid as these above properties are equivalent to the problem you posed.