Tom and Fred are inseparable twins. They are in a five-person club. How many ways can these five line up for a photo if Tom and Fred must stand together?
treat Tom and Fred as one unit, so there are only 4 to arrange
That would be 4!
But within each those, Tom and Fred could be switched.
So the total number of ways is 2(4!) = 48
To solve this problem, we can consider Tom and Fred as a single entity since they must always stand together. So essentially, we have four entities now: Tom & Fred (as one), plus the remaining three people in the club.
Now, we can arrange these four entities (Tom & Fred, Person 1, Person 2, and Person 3) in a linear order. The number of ways to arrange them can be found by considering Tom & Fred as a single entity and permuting the four entities.
Since we have four entities, there are 4! (read as "4 factorial") ways to arrange them. This means 4 x 3 x 2 x 1 = 24 possible arrangements.
However, within the Tom & Fred entity, they can arrange themselves in two ways (Tom on the left and Fred on the right, or Fred on the left and Tom on the right). So, we multiply the total number of arrangements by 2.
Therefore, there are 24 x 2 = 48 possible ways for the five-person club to line up for a photo if Tom and Fred must stand together.