Five combinatorics questions

Text

Answer the following questions, explaining the combinatorial formula you use.

(a) In how many ways can $6$ people be arranged around a table, if two arrangements are considered equal when every person has the same left and right neighbor?

(b) In how many ways can $20$ identical treats be distributed among $4$ hiding spots?

(c) In how many ways can $12$ different gifts be distributed among $5$ people (possibly all to the same person)?

(d) In how many ways can $12$ students be split into $3$ groups of $4$ (unordered groups)? Show that the answer is $5775$.

(e) In how many ways can $3$ Dobermans be chosen out of $12$ and $4$ pitbulls out of $7$?

Solution

(a) Circular permutations: fix one person as a reference point and arrange the remaining $5$ in the remaining seats: $(6-1)! = 120$.

(b) Combinations with repetition (stars and bars): distributing $20$ identical objects into $4$ containers amounts to choosing $3$ dividers among $23$ positions: $\binom{20+4-1}{3} = \binom{23}{3} = 1771$

(c) Each gift, being distinct, can go to any of the $5$ people independently of the others: $5^{12} = 244\,140\,625$

(d) Choose $4$ students out of $12$ for the first group, then $4$ out of the remaining $8$ for the second, and the last $4$ automatically form the third group; since the groups are unlabeled (unordered), divide by $3!$ to remove permutations of the same three groups: $\frac{1}{3!}\binom{12}{4}\binom{8}{4}\binom{4}{4} = \frac{495\cdot 70\cdot 1}{6} = 5775$

(e) The two choices are independent, so the combinations are multiplied: $\binom{12}{3}\binom{7}{4} = 220\cdot 35 = 7700$

$\boxed{120;\ 1771;\ 244\,140\,625;\ 5775;\ 7700}$