A has $3$ maps and B has $9$ maps. All the $12$ maps being distinct. Determine the number of ways in which they can exchange their maps if each keeps his initial number of maps.
My attempt is as follows:-
Suppose A has maps $M_1,M_2,M_3$ and B has maps $M_4,M_5,M_6,M_7,M_8,M_9,M_{10},M_{11},M_{12}$
Suppose A exchanges map $M_1$ with any of the $9$ maps B has, so no of ways = $\displaystyle{9\choose 1}=9$
After that he will exchange $M_2$ with any of the $9$ maps B had excluding the one which he already had i.e $M_1$, so total no of ways = $\displaystyle{8\choose 1}=9$
After that he will exchange $M_2$ with any of the $9$ maps B had excluding the one which he already had i.e $M_1$, so total no of ways = $\displaystyle{8\choose 1}=8$
After that he will exchange $M_3$ with any of the $9$ maps B had excluding the ones which he already had i.e $M_1,M_2$, so total no of ways = $\displaystyle{7\choose 1}=7$
So total no of ways should have been $9\cdot8\cdot7=504$, but actual answer is $219$
I also considered the case if question is not interested in the order of exchange rather only in the set of maps exchanged, in that case no of ways to exchange all $3$ maps with $9$ maps will be $\displaystyle{9\choose 3}=84$
I also considered the case if it is not necessary to exchange all maps. So I made following cases:-
Case $1$: Exchanging one map $\implies \displaystyle{3\choose 1}{9\choose 1}=27$ ways
Case $2$: Exchanging two maps $\implies \displaystyle{3\choose 2}{9\choose 2}=108$ ways
Case $3$: Exchanging all three maps $\implies \displaystyle{3\choose 3}{9\choose 3}=84$ ways
So finally I got the correct answer, how would one know that they are not talking about exchanging all maps rather atleast one?
As already stated in a comment, the question is indeed badly phrased.
An indication that “their maps” is not, as one might think, intended to imply that each person exchanges all their maps is that this is not possible, as one of them has more maps than the other. Since the inference that $B$ exchanges all her maps is clearly not warranted, one might consider also not drawing the corresponding inference that $A$ exchanges all his maps.