Hi, I need your help. I just cant figure out how to solve this problem.
Question:
There are two rows of chairs. The first row has two chairs and the second row has four chairs.
In how many ways can six people be seated if a couple from the group must sit in the same row?
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Help wil be appreciated. Thank you so much.
couple in 1st row
... 2 ways for 1st row
... 4! ways for 2nd row
couple in 2nd row
... 4*5 ways for 1st row
... 4! ways for 2nd row
[(4 * 5) + 2] * 4!
To solve this problem, we can consider the two cases separately:
Case 1: The couple sits in the first row.
In this case, we can treat the couple as a single entity. So, we have 5 remaining people to seat in the first row and the second row. We can arrange these 5 people in the first row in 5! (5 factorial) ways. Similarly, we can arrange the remaining person in the second row in 1 way. Therefore, there are 5! x 1 = 120 possible arrangements in this case.
Case 2: The couple sits in the second row.
Again, we treat the couple as a single entity, leaving us with 5 remaining people to seat. We can arrange these 5 people in the second row in 5! ways. Similarly, we can arrange the remaining person in the first row in 1 way. Hence, there are 5! x 1 = 120 possible arrangements in this case as well.
To get the total number of arrangements, we need to sum up the possibilities from both cases. Therefore, the total number of ways six people can be seated, given that a couple must sit in the same row, is 120 + 120 = 240.
In conclusion, there are 240 possible ways for the six people to be seated if a couple from the group must sit in the same row.