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.
 A: 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.
A: All $3$ boys have to be in different groups. Then we have $3$ choices which boy is in a group of $4$ and $\binom7{3,2,2}=210$ ways to choose girls for the groups (assuming the groups are unlabelled), for a total of $3\cdot210=630$.
A: Since groups are unlabeled 3 boys can be distributed only in 1 way and as 7 girls to be distributed as 2,2,3 into groups hence only 2 of the group be treated like identical groups.
= 630
For more similar to this there is question in other forum link is https://www.mathsdiscussion.com/forum/topic/permutations-and-combinations-2/?part=1#postid-73
A: 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$
