For the first problem, we would have $\binom{10+4-1}{4-1} = \binom{13}{3}$ ways of distributing the tourists.
Explanation:
This is an instance of the "stars and bars" problem. See here for details.
For the second problem, the answer would be $\binom{9}{3}$ as you want to choose places for 3 separators from amongst 9 spaces between the tourists. Details also here.
For the third problem, for the fully general case, you want to cast it as a recursion. See here for details. Since we are working with a small enough problem, we can break it into cases.
Case 1: Observe that if any 1 boat has 1 passenger, then the other three are restricted to have 3 passengers each. Since the boats are distinct, we choose the boat with 1 passenger in 4 ways.
Case 2: If a boat has a minimum of 2 and maximum of 3 passengers, then the only allocations are $\{2,2,3,3\}$ amongst the 4 ships. We can choose the 2 ships with 2 passengers in $\binom{4}{2}=6$ ways.
Thus the total number of allocations for the third problem is 4+6 = 10
For the fourth problem, we have to presume the kids are distinguishable from tourists? This is not very clear to me, and therefore I'm not sure. Please check in case the below makes sense:
Possible Approach: First divide 8 indistinguishable adult tourists amongst 4 ships with minimum of 1 and max of 3. Then allocate the kids separately keeping in mind constraint of maximum of 3.
Calculations: The ways of accocating 8 adults amongst 4 ships are $\left\{\{1,1,3,3\},\{1,2,2,3\},\{2,2,2,2\}\right\}$. We will consider these as cases
Case 1 - $\{1,1,3,3\}$: We will choose 2 boats to contain the 1 passenger each in $\binom{4}{2}=6$ ways. We can then allocate 2 kids to the first boat or second boat containing 1 passenger, or allocate them 1 each to the two boats - a total of 3 allocations. Thus the total number of ways is $6\times 3=18$
Case 2 - $\{1,2,2,3\}$: We will choose 2 boats to contain the 1 passenger in $\binom{4}{1}=4$ ways, and of the remaining boats, the one containing 3 passengers (of the 3 remaining ships) in $\binom{3}{1}=3$ ways. Therefore total number of ways of choosing adults in ships is $12$. We can then allocate (1) both kinds to the boat containing 1 passenger, (2) 1 to the boat containing 1 passenger and 1 to one of the boats with 2 passengers in 2 ways (3) allocate one kid each to the boat with 2 passengers. This is a total of 4 ways. Thus the total number of allocations is $12\times 4=48$
Case 3 - $\{2,2,2,2\}$: All boats contain only 2 passengers. We will choose the two boats to allocate 2 kids in $\binom{4}{2}=6$ ways.
Therefore the total number of allocations in all these cases is $18+48+6=72$ allocations for this question.
