Possible ways between two families to marry One family consists of 3 sons & 4 daughters.
Another family consists of 5 sons & 6 daughters .
In how many ways can a marriage be arranged between two families, provided that the youngest son or daughter of either family does not marry the eldest daughter or son of the other?
I have tried that for 6 months and I can't find a solution. Can somebody tell me how to work out please?
 A: Let us call the family with 3 sons and 4 daughters, A, and the other family, B.
I take it that marriages are "different-gender" marriages, and that all of family A get married.
Let us consider marriages between the 3 sons of A and the 6 daughters of B.
Total possible marriages $= 6\cdot5\cdot4 = 120$,
but these will contain combos where one or both of the symmetrical constraints are violated.
If, e.g. the eldest son marries the youngest daughter, violating one symmetrical constraint,
there are only $5\cdot4 = 20$ choices left for the remaining,
and if both constraints are violated, the only son left has only $4$ choices.
So we apply inclusion-exclusion to arrive at permissible combos.
$N = 6\cdot5\cdot4 - 5\cdot4 -5\cdot 4 + 4 = 84$
You can now work out permissible combos similarly for 4 daughters of A and 5 sons of B.
Finally, multiply the two figures. 
A: The sons of the first family can be paired with a daughter of the second family in $3\cdot6=18$ ways, but two of these pairings are forbidden. The daughters of the first family can be paired with a son of the second family in $4\cdot5=20$ ways, but two of these pairings are forbidden. Therefore the number of admissble pairings is $34$.
A: The procedure is separated in two independant events:
1-Marrying 3 sons to 6 daughters:
1.1-Marrying 4 middler daughters to 3 sons: 
$\binom34*3!$
1.2-Marrying 2 middler daughters and the oldest/youngest one to 2 mid sons + oldest/youngest:
$\binom24*2*2!$ for the youngest, and same value for the oldest. it means doubled.
2-Marrying 4 daughters to 5 sons:
1.1-Marrying 3 middler sons to 4 daughters: 
$\binom34*3!$
1.2-Marrying 2 middler sons and the oldest/youngest one to 2 mid daughters + oldest/youngest:
$\binom23*3*3!$ (x $2$)
The global ways of organizing unique instances of brides is (1)*(2)=
$$[\binom34*3!+2*\binom24*2*2!]*[\binom34*3!+2*\binom23*3*3!]$$
