Splitting $3$ boys and $7$ girls into 3 groups with restrictions There are $ 3 $ boys and $ 7 $ girls. How many ways can we divide them
  into three unlabeled groups such that each group has a boy, two groups have
  three people, and the third group has four people?


I think that the answer is $\frac{3!\times \binom{7}{2} \times \binom{5}{2}}{2}=630$, but other people said that it is $\frac{3\times \binom{7}{2} \times \binom{5}{2}}{3}=210$.

Edit: I did wrote something incorrect.
You're just identifying which of the girls go along with each boy.

So there are 3 ways to choose which of the boys is in the bigger group, and then $\binom73$ ways to choose whose his partners are.  Then there are $\binom42$ ways to choose which of the remaining girls goes with the older of the remaining boys.  Thus, the total is $3\cdot\binom73\binom42=3\cdot35\cdot6=630$.



The reason we don't need to do any dividing is because every grouping we are making are distinct.  If Adam, Ben, and Carl are the three boys, then putting Ben with Jane and Kim and Carl with Laura and Mary is a different grouping than if put Ben with Laura and Mary and Carl with Jane and Kim.  Each of the three groups CAN be distinguished -- by which boy is in the group -- so there is no need to worry about overcounting.
Choose 1 boy from 3, the 2 girls from 7 and 2 girls from 5 and then 3 girls from the 3 left.

So, total combinations: $ ^3C_1 \times ^7C_2 \times ^5C_2 =630$