Answer to Question #247646 in Discrete Mathematics for Anirudh

Question

There are n sisters in a family. They each have one dress. They decide to exchange their dresses such that nobody wears her own dress. They try all such combinations and take pictures. All the sisters appear in each picture and the order in which they appear does not count. Each possible way of dressing is photographed exactly once. If there are 44 such pictures, what is the value of n ?

Answer 1

Since there are $n$ sisters then, there are $n$ dresses also.

The dress for the $n^{th}$ girl can be won in $(n-1)$ ways and is given as,

$n(n-1)ways$.

The dress for the $(n-1)^{th}$ girl can be won in $(n-2)ways$

The dress for the$(n-2)^{nd}$ girl can be won in$(n-3)ways.$

This continues up to the dress of the $2^{nd}$ girl which can be won in $(n-(n-1))=1way$.

This number of ways can be written as,

$n(n-1)+(n-1)(n-2)+(n-2)*(n-3)+...+2(1)$.

Also, since each girl can wear $(n-1)$ dresses, we add this to get the total number of ways for taking the photographs.

Now, total number of ways to take photographs is,

$n(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+2+(n-1)=44...............(i)$

To find the value of $n,$ we proceed as follows,

When n=1, equation $(i)$ above gives, ways for the taking pictures,

When n=2, equation$(i)$ gives,$2(1)+(2-1)=2+1=3$ ways for taking pictures.

When n=3, equation$(i)$ gives,$3(2)+2(1)+(3-1)=6+2+2=10$ ways for taking pictures.

When n=4, equation$(i)$ gives,$4(3)+3(2)+2(1)+(4-1)=12+6+2+3=23$ ways for taking pictures.

When n=5, equation$(i)$ gives, $5(4)+4(3)+3(2)+2(1)+(5-1)=20+12+6+2+4=44$

The equation$(i)$ is satisfied when $n=5$.

Therefore, there are $n=5$ sisters in this family.