# Restricted Permutations and Combinations

Tino, Colin, Candice, Derek, Esther, Mary and Ronald are famous artist. Starting next week, they will take turns to display their work and each artist's work will be on display at the London Show for exactly one week so that the display of the artworks will last the next seven weeks. In how many ways can a display schedule be developed if

$(a)$. There is no restriction. ANS 7! EASY

Suppose that a revised timetable has been drawn up and the artists in $(a)$ are to display their work in groups of twos or threes so that the entire exercise takes at most $3$ weeks. In how many ways can Colin and Candice find themselves in the same group? Hint: We need to explore all possible scenarios - the possible distributions of the teams in terms of numbers are: $2,2,3; 2,3,2;$ and $3,2,2.$ Candice and Colin can be in the $1st, 2nd$ or $3rd$ week. This thinking gives us $9$ possible scenarios in which Adam and Brian may end up being in the same team. Answer: $150$

How to get the $150$ I have scratched my head to no avail.

What I assumed is that if they are both in a group that has only two people, we count that group as one person and if they are in a group that has $3$ people we count the group as two people.

I tried: In both cases the $n$ is reduced to $7-2$, and $r$ to either $3-2$ or $2-2$. Hence: $(5C1*2!*3!*2! + 5C0*2!*2!*2!) * 3!$

and many other futile attempts.

The problem is : 1). Classification, how do I classify this problem? As a strictly permutation or strictly combination or mixed permutation or combination?

2). Is sampling with or without replacement?

3). How do I change $n$ and $r$, by subtracting one from each? Or by subtracting p, the number of objects that will always occur? $7-2=5?$ or $7-1 =6?$

The correct reasoning approach will be greatly appreciated.

Where do you get $2*6!$ for (a)? I find $7!$
For the three week problem, let us start by assuming $3,2,2$. We can multiply by $3$ at the end to take care of cyclic permutations of weeks. The two pairs are differently named bins in this case. The two C's can be together in one of the twos in $2\text{(which week)}*{5 \choose 3}\text{(who is in the triplet)}=20$ ways. They can be two of the triplet in $5\text{(the other member of the triplet)}*{4 \choose 2}\text{(the first other couple)}=30$ ways. Multiplying by $3$ gives a total of $150$ ways.