Ten people line up for a photo, including you and your friend. How many line arrangements are there if the photographer insists on keeping 3 people in between you and your friend?
Hello, RTC1996!
Ten people line up for a photo, including you and your friend.
How many line arrangements are there if the photographer insists
on keeping exactly 3 people in between you and your friend?
There are 10 people: . $\displaystyle A\text{ (you)},\;B\text{ (your friend}),\:\text{ and 8 Others.}$
You and your friend are spaced like this:
. . $\displaystyle A\:\_\:\_\:\_\:B\;\;\text{ or }\;\;B\:\_\:\_\:\_\:A $ . . 2 ways
Select 3 of the Others and arrange them in the spaces: .
. . There are: .$\displaystyle _8P_3 \:=\:\dfrac{8!}{5!} \:=\:8\cdot7\cdot6 \:=\:336$ ways.
Now we have 6 "people" to arrange: .$\displaystyle \boxed{AxyzB}\; \text{and 5 Others}$
. . There are: .$\displaystyle 6! \,=\,720$ ways.
Therefore, there are: .$\displaystyle 2\cdot 336\cdot720 \:=\:483,\!840$ .line arrangements.
First, this is really nothing to do with probability.
Oh yes, it could be a question from a probability course.
But it is not probability: it is simply a counting question.
Once we select the three people to stand between you and your friend, which forms a block of ONE. Now in affect we have six blocks
Your original question asked to count the number of possible lines for the given restriction.
You could calculate the probability of 3 people being between you and your friend
if the 10 people line up randomly.
Nothing like that is being asked.
You are asked to count arrangements.
You and your friend can be in 6 different relative positions
YxxxFxxxxx
xYxxxFxxxx
xxYxxxFxxx
xxxYxxxFxx
xxxxYxxxFx
xxxxxYxxxF
now you can swop places with your friend.
All the other people can move about in the "x" positions.
Soroban split it up to count them easily.
Think of the "rectangle" as a "person" or single group moving in unison.