First question, $n$ families , each family has a father, mother, child and a dog, sit around a circular table.
in how many ways can they be seated :
- no conditions
- every child sits between the parents
- every couple sits together, and no dogs are next to each other.
Second question, in how many ways can we place $4$ objects with different weights in $7$ drawers that are sorted one above the other (they are distinguishable) in the following conditions:
- no conditions
- the lightest and heaviest objects are not together
- the highest drawer has exactly one object and the lowest drawer has at least 1 object
- each drawer has at most one object
First question:
for the first part we have $n$ families , $1$ circular table and each family has $4$ members , so we have in total $4n$ seats and according to circular permutation formula it will be $\frac{(4n)!}{4n}$
the second part we first seat one of the parents (doesn't matter who) in a certain place around the table and then we seat the child we have 2 ways to seat the parents 2 ways to seat the child and 2 ways to seat the dog so $2\cdot2\cdot2=8$ and we need to do this for $n$ families so $8^n$
I did not understand the last part does that mean the parents are considered now $1$ instead of $2$ or does it mean that they sit next to each other?
Second question:
first part: $P(7,4)=7 \cdot 6 \cdot 5 \cdot 4 = 840$ sorting 4 different objects in 7 different drawers
second part: assume the objects are $\{a,b,c,d\}$ with the lightest being $a$ and heaviest $b$
first lets put $\{a,b,c\}$ in the 7 drawers $P(7,3)=7 \cdot 6 \cdot 5 =210$
same thing for $\{b,c,d\}$ but $a$ must be in one drawer so $P(6,3)=6 \cdot 5 \cdot 4 =80$ lastly we sum the options and get $210+80=290$
third part: let $\{A,B,C,D,E,F,G\}$ be the drawers while $A$ is the one on the bottom and $G$ being the highest we will put one object in $G$ , we have $4$ ways of doing that so we are left with $3$ objects and $6$ drawers while the bottom one has at least $1$ object so we have 3 cases
- $3 \choose 1$ $\cdot P(5,2)=60$
- $3 \choose 2$ $\cdot P(5,1)=15$
- $3 \choose 3$ $\cdot P(5,0)=1$
summing up all the options we get $4+60+15+1=80$
fourth part: the first object has $7$ options the second has $6$ third has $5$ and fourth has $4$ so we get $7+6+5+4=22$
are my ways correct? these are the only questions in the book without answers.. thanks for any tips and help hopefully my translations are understandable
Best Answer
First question:
Part $1$: There are $4n$ distinguishable "sitters." Assuming that two arrangements are the same if they differ only by a rotation, since there are $4n$ possible rotations for each permutation, there are $\dfrac{(4n)!}{4n}=(4n-1)!$ possible arrangements. Your answer is correct but can be expressed more simply.
Part $2$: If every child sits between the parents, then consider parent-child-parent as a single unit. There are, therefore, $2n$ units to arrange, which can be done in $(2n)!$ ways. However, each parent-child-parent unit can be formed in $2$ different ways, so there are $2^n \cdot (2n)!$ possible permutations. Again we must divide by $4n$ to account for rotations, so there are $2^{n-1} \cdot (2n-1)!$ possible arrangements. Your answer is not correct because the problem doesn't require a family's dog to be adjacent to the family.
Part $3$: If every couple sits together and none of the dogs sit next to each other, first arrange the couples and the children. There are $2n$ units to arrange ($n$ pairs of parents and $n$ children), which can be done in $(2n)!$ ways. There are $2^n$ ways to arrange the parents within each pair.
There are $2n$ "gaps" into which we can place the dogs. The first dog can choose any of the $2n$ gaps. The second dog has $2n-1$ gaps to choose from, and so on. We can therefore place the dogs in $\dfrac {(2n)!}{n!}$ ways. Another way to see this is that there are $\binom {2n}{n}$ ways to choose the gaps the dogs will fill, and once that's done there are $n!$ ways to arrange the dogs within those gaps, so there are $n! \cdot \binom{2n}{n}=\dfrac{(2n)!}{n!}$ ways to place the dogs.
Thus, there are $2^n \cdot (2n)! \dfrac{(2n)!}{n!}$ possible permutations. Once again we need to divide by $4n$ to account for rotations, yielding $2^{n-1} \cdot (2n-1)! \dfrac{(2n)!}{n!}$ possible arrangements.
Second question:
Part $1$: If any object can go into any drawer, then each of the $4$ objects has $7$ possible locations, so there are $7^4=2401$ possible arrangements. Your answer is incorrect because it's possible for a drawer to contain more than $1$ object.
Part $2$: If the lightest and heaviest object must go into different drawers, then (going in ascending order by weight) there are $7$ choices for each of the first $3$ objects, but only $6$ choices for the $4$th object, so there are $6 \cdot 7^3=2058$ possible arrangements. Your answer is incorrect because you "inherited" the mistake you made in Part $1$.
Part $3$: If the highest drawer has exactly $1$ object, there are $4$ choices for what that object can be. There are $6^3=216$ ways to arrange the other $3$ objects among the remaining $6$ drawers without constraint. However, in $5^3=125$ of those arrangements, the lowest drawer has no object. Therefore, there are $4 \cdot (216-125)=364$ permissible arrangements. I'm not sure how you got the various figures you're combining but your answer is not correct.
Part $4$: If each drawer has at most one object, then there are $7$ choices for the first object, $6$ choices for the second object, $5$ choices for the third object, and $4$ choices for the fourth object, for a total of $7 \cdot 6 \cdot 5 \cdot 4=840$ possible arrangements. Your answer is not correct because you added when you needed to multiply.