How many $4$-digit numbers with non-repeating digits can be written by choosing $2$ digits from $A=\{1,2,3,4\}$ and $B=\{3,5,6,7\}$

Question

For $A = \{1, 2, 3, 4\}$ and $B = \{3, 5, 6, 7\}$

How many different $4$-digit numbers with non-repeating digits can be written by choosing $2$ digits from $A$ and $2$ digits from $B$?

So I've tried solving this problem by doing:

$${4 \choose 2}{3 \choose 2}4!$$

Choosing $2$ out of the $4$ possible numbers from $A$, then choosing another $2$ out of the $3$ possible from $B$ (not counting $3$ so it's not repeated) then multiply it by $4!$ to find all the different arrangements. I found $432$ but the answer is $648$. Why didn't my method work? Which $216$ numbers did my answer not include, and why? Thanks!

Answer 1

There are two cases: $3$ is taken or not taken from $A$: $$\left[{3\choose 1}{3\choose 2}+{3\choose 2}{4\choose 2}\right]\cdot 4!=648.$$ When you consider ${4\choose 2}$ from $A$, it can contain $3$ or not contain $3$. When it does not contain, you can take $3$ from $B$, which you are missing.

Answer 2

Another way to think about it is, first choose $2$ from each set have $\binom42^2$,

multiply $4!$ to find all the different arrangements,

then take out the repeated percentage $\left(1-\frac{1}{4}\right)$ we get:

$$4!\binom42^2\left(1-\frac{1}{4}\right)=648$$

Answer 3

i think that the answer is $${4 \choose 2}{4 \choose 2}4!-{3 \choose 1}{3 \choose 1}4!$$ you calculating all the possible solutions and theν you subtract the solution that doesnt work for you ( from A choose 3 AND from B chosse 3)