Three couples are seated in two rows of three seats. If no row or colu

Question

Three couples are seated in two rows of three seats. If no row or column of seats contains both members of a couple, how many possible arrangements are there?

A. 48
B. 96
C. 128
D. 192
E. 288

Bismarck · Accepted Answer

OA:B Let three couple be {A1,A2};{B1,B2};{C1,C2} There are two patterns of seating possible to meet the conditions required by the question stem. veritas prep.PNG Explaining calculation for pattern 1 For Seat 1, there are 6 possibilities (3*2 people). Whoever is selected for seat 1, his/her partner will sit on seat 5 For Seat 2, there are 4 possibilities (3*2 people -1*2 people already seated). Whoever is selected for seat 2, his/her partner will sit on seat 6. For Seat 3, there are 2 possibilities (3*2 people -2*2 people already seated). Whoever is selected for seat 3, his/her partner will sit on seat 4. Total Arrangement in pattern 1: 6*4*2 =48 Calculation for pattern 2 can be done in same way Total Arrangement in pattern 2: 6*4*2 =48 Final number of arrangement = 48+48 =96

globaldesi · Answer

For any permutation with multiple restrictions, it is often easier to consider how each position can be picked (the slotting method) rather than trying to figure out how to manipulate the permutation equation to match the situation given. This is an example of one of these situations – it is easy to imagine different acceptable permutations but difficult to see what that might mean to a generalized permutation formula.

Think of this instead as six “slots” that must be filled and use basic counting principles. Consider how many ways each can be filled, then multiply the different possibilities together.

The first seat can be filled by any of the six people, since there are not yet any restrictions.

The second seat can be filled by anyone except the other half of the first couple. This means it can be filled 4 ways.

The third seat can then be filled by either of the two members of the last couple.

To find the total number of ways the first row can be arranged, just multiply these three numbers together to get:

6 x 4 x 2 = 48

As you start the second row, notice that there are only three people left. The first “slot” of the second row can be filled by anyone except the other half of the couple in the first seat in the first row. This means that there are two possibilities as to how to fill this chair.

There is only one way to fill each of the remaining chairs. To see why, consider an example. If the three couples are just represented as Aa, Bb, and Cc, look at what happens if the first row is in the order ABC and the second row starts with b:

A B C
b _ _

The only possible place to put ‘c’ is in that second position if you want to avoid a column from having two members of the same couple. There is only 1 way to choose the second two seats and 2 total ways to pick the second row.

The total number of arrangements can then be found by multiplying the value found for the first row by the value found for the second row to get 48 x 2 = 96.

MarcoAD · Answer

First, let's calculate how many arrangements can we have on a single row of 3 seats.

6!/3!=120

Now, let's remove from the total all the arrangements we don't like, i.e. both member of one couple seating in the same row. Calling the couples like A1 & A2, B1 & B2, C1 & C2, we have:

A1,A2,X
A1,X,A2
X,A1,A2
X,A2,A1
A2,X,A1
A2,A1,X

3!=6 arrangements

X can be either B1, B2, C1 or C2, meaning that we have 6*4=24 unwanted arrangements for what concern couple A.

Doing the same for couple B and couple C, we have a total of 24*3=72 unwanted arrangements. Or, in other words, 120-72=48 arrangements on the first row which can be acceptable.

Now let's look at the second row. The number of acceptable arrangements here it's again 48 but this time we have to look who is seating in the first row! If 3 members are seating in the first row, the other 3 must seat in the second, this already eliminate many arrangements from the initial 48. Let's assume in the first row A1, B1 and C1 are seating in the order just written; it turns that in the second row A2, B2 and C2 are seating and all we have to do is calculate the acceptable arrangements of only these 3 members. Jotting down the possibilities, it turns that there are only 2 acceptable arrangements:

A1 B1 C1 
B2 C2 A2

&

A1 B1 C1
C2 A2 B2

You can easily verify that this applies for each arrangements on the first row.

Hence, the answer is 48*2=96, option B.

TheNightKing · Answer

So I did not solve it all the way but took a calculated guess.
So the first thing, first person can sit anywhere. which means 6 ways. Answer has to be multiple of 6. Unfortunately only C is gone.
When you try to make next few people sit you realize you have restricted options available (only 2). So when you realize the answer wont be as big as 192. So you are down to A and B. I selected B and it was the right choice.

Now, when you solve it all the way you work out possible arrangements are 6*2*2*2*2=96. Hence B

Akshat1994 · Answer

Say 3 couples be AB, CD, EF

A can sit in 6C1 ways, correspondingly B can sit in 2 ways.

Now C can sit in 4C1 ways, correspondingly D can sit in one way.

E can sit in 2 ways, correspondingly F in one way

Total arrangements= 6C1*2*4C1*1*2C1*1= 96

Posted from my mobile device

bumpbot · Answer

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Three couples are seated in two rows of three seats. If no row or colu

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Three couples are seated in two rows of three seats. If no row or colu

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