number of images from $A$ to $B$ Let $A = \\{2,4,6,8,10,12\\}$ and $B = \\{3,7,11\\}$. Then no. of function such that element $3$ of $B$ has two , $7$ of $B$ has three and $11$ of $B$ has one pre - image in $B$ ,is My Try:: for element $3$ of $B$. we have to choose two element in $A$ out of total six elements. This can be done by $\displaystyle \binom{6}{2} = 15 \;\; $ways Similarly for element $7$ of $B$. we have to choose three element in $A$ out of total four elements. This can be done by $\displaystyle \binom{4}{3} = 4 \;\; $ways Similarly for element $11$ of $B$. we have to choose one element in $A$ out of total one elements. This can be done by $\displaystyle \binom{1}{1} = 1 \;\; $ways So Total ways $ = 15 \times 4 \times \times 1 = 60$ can anyone explain me i have done right or not if not then how can i proceed Thanks

Quoth André Nicolas:

Yes, you are right. In writing up this sort of thing, it would be good if you said that for every choice of preimage for $3$, there are $4\choose 3$$\ldots$ to make the logic clear. And maybe at the end say that now the preimage of $11$ is determined, multiplying by $1\choose 1$ seems mechanical.