In how many ways can 6 students be seated in a row of 6 seats if 2 of the students insists o sitting beside each other?
treat the 2 as one item, so you are arranging 5
which is 5!
but the two students could be interchanged, so
number of ways to seat the students
= 2(5!) = ...
65
To solve this problem, we can treat the pair of students who want to sit together as a single entity. So, instead of having 6 students, we now have 5 entities (one of them being the pair) to arrange in a row of 6 seats.
First, we can arrange the 5 entities in 5! (5 factorial) ways. This means there are 5 × 4 × 3 × 2 × 1 = 120 possible arrangements for the 5 entities.
Within the pair themselves, there are 2! (2 factorial) ways to arrange the students. This means there are 2 × 1 = 2 possible arrangements for the pair.
To calculate the total number of arrangements for the 6 students, we multiply the number of arrangements for the 5 entities by the number of arrangements for the pair, which gives us:
Total arrangements = 5! × 2!
= 120 × 2
= 240
Therefore, there are a total of 240 ways to seat the 6 students in a row of 6 seats if 2 of them insist on sitting beside each other.