is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

You will get a detailed answer to your question or assignment in the shortest time possible.

## Here's the Solution to this Question

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.