combinatorics,choose specific number of elements from sets I was just wondering...
If we had 4 sets ,lets say :
\begin{align}
S_1 &= \{a,b,c,d\}\\
S_2 &= \{a,e,f,d\}\\
S_3 &= \{a,b,f\}\\
S_4 &= \{b,g\}
\end{align}
In how many ways can we choose two $a$'s ,one $f$ and one $b$ ?
I am stuck with this problem and would be glad to hear your ideas.
There must be chosen only one element from a 
set !
 A: If the sets elements are considered to be different and as they can be used once then its $3.2.{3 \choose 1}=18$
A: I assume that you have to pick one element from each set?!
It's easier than you think.
The $b$ has to come from $S_4$, as there is no $g$.
Then at least one $a$ from $S_1$, as there are no $b,c,d$ left to choose.
Then we have one $a$ and one $f$ left that can come from $S_2$ and $S_3$ and the other way around, so I'd say $2$ ways.
A: if we consider each set as an element a because each set contains only one a,we have 3C2 possible ways to choose 2 a's ,
similarly 2C1 for 1 f and 3C1 for 1 b
           That's how we have total number of 3C2*2C1*3C1 possible number of ways i.e. 18

A: Here’s a more tedious, but more easily generalizable approach.
You can enumerate all the results of choosing one element from each set by evaluating the polynomial product $(a + b + c + d) (a + e + f + d) (a + b + f) (b + g)$. The number of ways to choose two $a$’s, one $f$, and one $b$ is the coefficient of $a^2f^1b^1$ (or $a^2bf$) in the result. This is the same as the coefficient of $a^2bf$ in the same product after setting $c$, $d$, $e$, and $g$ to zero, that is, in the product $(a+b)(a+f)(a+b+f)b$. This is not so hard to expand, and it equals $a^3 b+2 a^2 b^2+\color{red}{2 a^2 b f}+a b^3+3 a b^2 f+a b f^2+b^3 f+b^2 f^2$, so the answer is 2.
Or, for this specific problem, when you go to look for the coefficient of $a^2bf$ in $(a+b)(a+f)(a+b+f)b$, you can see from the last factor $b$ that you need the coefficient of $a^2f$ in the product $(a+b)(a+f)(a+b+f)$, and can set $b=0$ there. Now you want for the coefficient of $a^2f$ in the product $a(a+f)^2$, or equivalently, the coefficient of $af$ in $(a+f)^2$, which is again 2.
