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## Difficult Permutation Problem

I'm having a lot of trouble making my answer look like the OA. Can someone help?

Here's the problem:

In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?

OA to follow after some discussion

2. Good post? |
Originally Posted by choked
I'm having a lot of trouble making my answer look like the OA. Can someone help?

Here's the problem:

In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?

OA to follow after some discussion
It depends on whether or the each seat at a table is unique. What I mean is that if we consider the tables to have chair1, chair2, chair3, etc... we get a different answer than if we just think of the tables as having 5 or 7 chairs. In other words, does seating the people in the same order but shifting everyone over to the right or the left count as the same arrangement or not?

If we're only concerned about which people are next to each other and not which particular chairs they occupy, then we have 12c5 (or 12c7) ways to divide them into groups of 7 and 5. Then we have 6! ways to arrange the people at the larger table and 4! ways to arrange the people at the smaller table.

This gives us (12c5)*6!*4!, which will be a huge number.

If we care about which chairs they occupy, then the 6! and 4! become 5! and 7!. They would cancel out with the denominator in 12c5, so we would just get 12!.

These calculations seem right to me for these 2 different cases. What do you think?

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There could be an umpteen combination:
12C5 + 12C7 = calculate it, that is onerous! But what does the problem requires? let 7 people sit across first table and the left 5 people sit across other table? Or otherwise?

4. Good post? |
Originally Posted by choked
I'm having a lot of trouble making my answer look like the OA. Can someone help?

Here's the problem:

In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?

OA to follow after some discussion
You can choose 7 people from 12 in 12C2 ways.
7 people can be seated in 7! ways
Remaining 5 can be seated in 5! ways

Total no of ways = 12C2 * 7! * 6!

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Since the question has made no mention of which types of tables these are (ie. a round table will have a different number of distinct seating arrangements than a rectangular one) or the order of how they're seated at the table, I'm assuming that this is just a simple combinations problem. Since every distinct group of 7 people corresponds to an equally distinct group of 5, the answer is simply 12C7 = 12C5 = 12*11*10*9*8/120 = 792

The only other answer that I would consider (albeit loosely based on the vague nature of the question) would be 2*792 = 1584 (assuming each table can hold either 7 or 5 people)

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## Re: Difficult Permutation Problem

In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?

Ans: The first sit can be taken by anyone of 12 people, the next sit by the remaining 11 people and so on. The total number of such arrangements is: 12!
In my humble opinion, it’s immaterial how we subdivide these arrangements into separate groups be that (5 people at the first table and 7 at the second) or otherwise, or say (6 people at the first table and 6 at the second) still the total number of such arrangements will not change.

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## Re: Difficult Permutation Problem

Originally Posted by Tristar
In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?

Ans: The first sit can be taken by anyone of 12 people, the next sit by the remaining 11 people and so on. The total number of such arrangements is: 12!
In my humble opinion, it’s immaterial how we subdivide these arrangements into separate groups be that (5 people at the first table and 7 at the second) or otherwise, or say (6 people at the first table and 6 at the second) still the total number of such arrangements will not change.
If the order of the individual seat locations matters, then you are correct: the answer is 12!

Note that this number can also be derived as follows:

# of Groups of 7 (and 5 left over) that can be created from a population of 12 = 12C7 = 12C5 = 12! / (7! * 5!)

# of unique arrangements of 7 people in 7 distinct chairs = 7!
# of unique arrangements of 5 people in 5 distinct chairs = 5!

# of total unique seating arrangements = 12C5 * 7! * 5! = 12!

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