Charlene wants to make a string of beads with 10 blue beads and two metal beads. How many different strings of this type can be made? Charlene has 21 metal beads and 12 blue beads.


*

*blue beads are same. 

*metal beads are different. 


Charlene wants to make a string of beads with 10 blue beads and two metal beads.
How many different strings of this type can be made?
I know how to do r permutation or r combination when objects are all different (eg all metal beads). for example only one type of beads.

 A: Let's start with the blue beads. The blue beads are identical so there is only one way to select ten blue beads. Another way to say this is no matter which blue beads you select you will end up with an identical looking string of ten blue beads so there is really only one final result. So let's place these ten beads on the string:
$$
\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_
$$
Notice that I placed spaces in between the beads. We will use these for placing the metal beads but first we must select the metal beads. Each of the $21$ metal beads are unique so there are ${21}\choose{2}$ ways to select two beads. Now that we have these two beads let's label them $M_1$ and $M_2$.
The last thing we have to do is place the metal beads. Let's start with $M_1$. Looking at the string of blue beads above we can see that there are $11$ slots where we can place $M_1$. As such there are ${{11}\choose{1}} = 11$ ways to place $M_1$. Here is one such way:
$$
\_B\_B\_B\_B\_B\_B\_B\_B\_M_1\_B\_B\_
$$
Note that I've kept spaces around $M_1$. We need this because $M_2$ could be placed directly on either side of $M_1$ as well as in any of the other "blue" spaces (this is true even if $M_1$ is at one of the ends but it is always good to consider these cases). Now we can see that there are $12$ slots where we can place $M_2$. As such there are ${{12}\choose{1}} = 12$ ways to place $M_2$. In summary we have
$$
\begin{align}
&\;\;\;\;1 &\mbox{ways to select 10 identical blue beads from 12} \\
&{{21}\choose{2}} &\mbox{ways to select 2 unique metal beads from 21} \\
&{{11}\choose{1}} &\mbox{ways to select a spot for } M_1 \\
&{{12}\choose{1}} &\mbox{ways to select a spot for } M_2
\end{align}
$$
Leaving us with a total of $1 \cdot {{21}\choose{2}} \cdot {{11}\choose{1}} \cdot {{12}\choose{1}}$ ways to select to string of beads. One thing you should clarify however is whether the string can be flipped or not. (i.e. Is $BM_1BM_2$ the same as $M_2BM_1B$?) If there are the same then you will need to divide to above result by $2$ (there aren't any "symmetric" strings so we've double counted everything).
