How many bracelets can be formed? If I am about to fabricate a bracelet and I can select $24$ pearls out of a total of $500$ pearls ($300$ white, $150$ red and $50$ green) how many possible bracelets can I create?
The (official) solution to this question is 
$$ \frac{500!}{476!} = 3.4\cdot10^{64} $$
But how can it be that the colors of the pearls were totally neglected? I mean if there would just be red pearls then this solution would be clear to me. But since there are different colors I am confused...
 A: The first answer is excellent. I would just like to remark that if we are calculating the cycle index of $D_{24}$ anyway, then why not use the Polya enumeration theorem (PET)?
We have by table lookup and/or basic reasoning that for the dihedral group the cycle index is given by
$$ Z(D_n) = \frac{1}{2} Z(C_n) + 
\frac{1}{4} \left(a_1^2 a_2^{(n-2)/2} + a_2^{n/2}\right)$$
and
$$ Z(C_n) = \frac{1}{n} \sum_{d|n} \phi(d) a_d^{n/d}.$$
This implies that
$$ Z(D_{24}) =
1/48\,{a_{{1}}}^{24}+{\frac {13}{48}}\,{a_{{2}}}^{12}+1/24\,{a_{{3}}}^
{8}+1/24\,{a_{{4}}}^{6}+1/24\,{a_{{6}}}^{4}+1/12\,{a_{{8}}}^{3}+1/12\,
{a_{{12}}}^{2}\\+1/6\,a_{{24}}+1/4\,{a_{{1}}}^{2}{a_{{2}}}^{11}.$$
The answer to the problem is then given by $Z(D_n)(x+y+z)_{x=1, y=1, z=1},$ but we have
$$ Z(D_n)(x+y+z) =
1/48\, \left( x+y+z \right) ^{24}+{\frac {13}{48}}\, \left( {x}^{2}+{y
}^{2}+{z}^{2} \right) ^{12}\\+1/24\, \left( {x}^{3}+{y}^{3}+{z}^{3}
 \right) ^{8}+1/24\, \left( {x}^{4}+{y}^{4}+{z}^{4} \right) ^{6}+1/24
\, \left( {x}^{6}+{y}^{6}+{z}^{6} \right) ^{4}\\+1/12\, \left( {x}^{8}+{
y}^{8}+{z}^{8} \right) ^{3}+1/12\, \left( {x}^{12}+{y}^{12}+{z}^{12}
 \right) ^{2}+1/6\,{x}^{24}+1/6\,{y}^{24}+1/6\,{z}^{24}\\+1/4\, \left( x
+y+z \right) ^{2} \left( {x}^{2}+{y}^{2}+{z}^{2} \right) ^{11}$$
so that $$Z(D_n)(x+y+z)_{x=1, y=1, z=1} = 5884491500$$
confirming the result from the first answer.
Since this problem gets asked quite frequently I am also posting the code for the solution using the PET in Maple for a necklace with any number of beads. Values on the order of $24$ with a lot of divisors usually require some assistance by a CAS. The routine pet_cycleind produces the cycle index and pet_varinto_cind does the substitution of the repertoire generating function.

with(numtheory);
with(group):
with(combinat):


pet_cycleind := 
proc(n)
        local d, s, t;

        s := 0;

        for d in divisors(n) do
            s := s + phi(d)*a[d]^(n/d);
        od;

        if type(n, odd) then
           t := n*a[1]*a[2]^((n-1)/2);
        else
           t := n/2*(a[1]^2*a[2]^((n-2)/2)+a[2]^(n/2));                
        fi;        

        (s+t)/2/n;
end;

pet_varinto_cind :=
proc(poly, ind)
           local subs1, subs2, polyvars, indvars, v, pot, res;

           res := ind;

           polyvars := indets(poly);
           indvars := indets(ind);

           for v in indvars do
               pot := op(1, v);

               subs1 := 
               [seq(polyvars[k]=polyvars[k]^pot, 
               k=1..nops(polyvars))];

               subs2 := [v=subs(subs1, poly)];

               res := subs(subs2, res);
           od;

           res;
end;

A: Mathematically, bracelets are defined inequivalent under rotations and reflections (i.e., the dihedral group action).  Moreover, e.g. one red bead should be indistinguishable from another red bead, otherwise we should just call them "labelled beads".
Under this interpretation, the fact that there are $300$, $150$ and $50$ beads of the various colours, essentially means we have an unlimited supply of each colour (since there's more than $24$ in each case).  In this context, the question should be written along the lines:

How many $24$-bead necklaces can be made with beads of $3$ colours?

The best method for answering this question is via Burnside's Lemma.
Let $L$ be the set of $24$-element sequences of $\{\text{white},\text{red},\text{green}\}$.  We see $|L|=3^{24}$.  The dihedral group $D_{48}$ acts on $L$ by permuting the indices of the elements.  To use Burnside's Lemma, we need to count how many necklaces are fixed by the action of $\alpha$, for each $\alpha \in D_{48}$.
If $\alpha \in D_{48}$, and $\alpha N=N$ for some $N \in L$ (i.e. $\alpha$ is an automorphism of $N$), then the group generated by $\alpha$, namely $\langle \alpha \rangle$, acts on the elements of $N$.  The beads in the same orbit under this action have the same colour (since $\alpha N=N$).  So there are $3^{\#\text{orbits}}$ necklaces that are fixed by any $\alpha \in D_{48}$ (we pick a bead colour for each orbit).  The number of orbits is the number of cycles in the cycle decomposition of $\alpha$.  Hence $\alpha$ fixes exactly $3^{\#\text{cycles of } \alpha}$ necklaces.
We can do some bookkeeping, e.g. with GAP, to obtain:
$$\begin{array}{c|cc}
\text{cycle structure} & \text{nr perms in } D_{48} \text{ with that cycle structure} & \text{nr cycles} \\
\hline
1^{24} & 1 & 24 \\
1^2 2^{11} & 12 & 13 \\
2^{12} & 13 & 12 \\
3^8 & 2 & 8 \\
4^6 & 2 & 6 \\
6^4 & 2 & 4 \\
8^3 & 4 & 3 \\
12^2 & 4 & 2 \\
24 & 8 & 1 \\
\end{array}$$
So, by Burnside's Lemma, the number of necklaces is $$\frac{1}{48} \left( 3^{24} + 12 \times 3^{13} + 13 \times 3^{12} + 2 \times 3^8 + 2 \times 3^6 + 2 \times 3^4 + 4 \times 3^3 + 4 \times 3^2 + 8 \times 3^1 \right)$$ which is $5884491500$.
A: Yeah it seems like the official answer is assuming each pearl is distinct, in which case the fact that it provided colors is meaningless.  On top of that even if you assume each pearl is distinct, the official answer is wrong.  It would be right if the question was asking about putting pearls in a line instead of a bracelet.  If the question were asking for (say) putting 24 pearls on a merry-go-round, then you need to divide the official answer by 24, because rotation doesn't matter anymore.  Now if it were a bracelet, take the merry-go-round answer and divide that by 2 (you can put on a bracelet 'forwards' or 'backwards').
So to sum up:  Assuming all pearls are distinct, picking 24 where order matters is:
$$ {500 \choose 24} = \frac{500!}{24! \cdot 476!}.$$
Where order matters, ie putting the pearls in a line (official answer):
$$ \frac{500!}{24! \cdot 476!} \cdot 24! = \frac{500!}{476!}.$$
Now if you were putting the pearls on a bracelet:
$$ \frac{500!}{476!}  \frac{1}{24 \cdot 2} .$$
Seems it's just a very poorly written question....
