Lakers and Bucks arrangements I know some of them must be trivial but I have pretty poor knowledge of combinatorics:
We have 5 Lakers players and 7 Bucks and we want to
(a) have them seated in a row. My answer: $(7+5)! = 12!$
(b) In a row but with all Lakers to be seated together: We consider the Lakers as one block, so we have 8 "items":  $8!5!$.
(c) Be seated in a circle: It is $(12-1)! = 11!$
(d) In a circle but with at least 2 Lakers players to be seated together: I am not sure about this one: We treat the pair as one, so we have $10!$?
(e) In a row, with two Lakers in the first and last position and no two Lakers in consecutive places. I don't know how to calculate this one. Clearly we need to find the arrangement of 10, with no Lakers in starting and ending positions, and no Lakers in consecutive places, multiplied by the number of ways to select 2 out of 5 (which is 20?)
(f) In a circle but with no two or more Lakers sitting next to each other. I don't know how to calculate this one
(g) In how many ways can we make a subgroup of any 3 players: It is $\binom {12}{12-3}$
(h) a subgroup of size 3 but with exactly 1 Laker player: $\binom {7}{2} \binom {5}{1}$
(i) a subgroup of size 3 but with at least 1 Laker player: $\binom {12}{12-3} - \binom {7}{3}$??
Can you please help me with the ones I've got wrong or haven't answered?
Thank you!
 A: I've checked a, b, c, g, h, i and they look correct.
Hints for the remaining ones:

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*(e) You are right about the 20 ways to select the two Lakers on the ends. There are $7!$ ways to arrange the Bucks players relative to each other, and $3!$ ways to arrange the remaining Lakers players relative to each other. It remains to count the number of ways to arrange these $7$ Bucks and $3$ Lakers. There are four "gaps" between/around the three Lakers, and there must be at least one Buck in each of these gaps. This is the number of compositions of $7$ into $4$ positive numbers. You can count this manually, i.e. $4+1+1+1$, $3+2+1+1$, $1+2+2+2$, etc.

*(f) A similar argument can be used here. There are $4!$ ways to arrange the Lakers in a circle. There are $5$ gaps between them that each must contain at least one Bucks player. You can manually count the number of compositions of $7$ into $5$ positive numbers; they are the five permutations of $2+1+1+1+1$. Once you have decided where the Bucks will seat, there are $7!$ ways to fill the seats.

*(d) This is the complement of (f) among all circular seatings

