I've found interesting problem in a puzzle book for 6-10

Question

I've found interesting problem in a puzzle book for 6-10 years old. Try to solve.

4 couples enter a restaurant. How many ways can they be seated at a round table so that the men and the women alternat and no husband and wife sit next to each other?

smandalika · Answer

tough one lemme take a shot at it.

First off to find out all possible combinations of man woman sitting alternately we pin down a man to chair.now among the rest of the men the total number of combinations possible is 3!. The number of ways the remaining woman can be seated is 4! ways. so the total ways possible is 4!*3!

let us calculate the the number of ways a husband wife pair will always be together.
that would be following the same rule of seating a couple down and calculating the combinations which would be 3!. but the couples can switch seats between themselves so the total combs will be 3!* 2 

hence the final answer(never togther ) will be 4!*3! - 3!*2 =132 ways

whats the OA

boksana · Answer

The OA I'll tell later. 
But the point is that 6-10 years old kids do not know all these combinations formulas and still can solve this problem. We have to think outside the box.

boksana · Answer

Hint: 6 years old kid might not know number 132, so your answer is wrong

smandalika · Answer

tough one lemme take a shot at it.

First off to find out all possible combinations of man woman sitting alternately we pin down a man to chair.now among the rest of the men the total number of combinations possible is 3!. The number of ways the remaining woman can be seated is 4! ways. so the total ways possible is 4!*3!

let us calculate the the number of ways a husband wife pair will always be together.
that would be following the same rule of seating a couple down and calculating the combinations which would be 3!. but the couples can switch seats between themselves so the total combs will be 3!* 2 

hence the final answer(never togther ) will be 4!*3! - 3!*2 =132 ways

whats the OA

ian7777 · Answer

great question (too hard for gmat, but still great).

Is it 60?

I have 4x4x3x3x2x2x1x1 for the 8 spaces around the table.
Then I divide that number by 8, since it's round, and each permutation would have 8 copies. So that's 72.

Then do it again by couples to put the couples together: 4x3x2x1, and divide that answer by 4, since, as I said above, it's round, but now there are 4 spaces. So that's 6. But then times by 2, since each couple could sit man/woman or woman/man. So that's 12.

72-12=60

jpv · Answer

Let me try my hands on this...

Let Husbands: H1 H2 H3 H4
Wives : W1 W2 W3 W4

On a round table fix one person say H1.
Now, rest all husbands can sit in 6 ways.
( H2 --> (h3, h4) (h4, h3))
( h3 --> (h2, h4) (h4, h2))
( h4 --> (h2, h3) (h3, h2))

What we mathematically say (3!)

Now for wives....
between h1 and h2 we can only arrange w3 and w4 (2!).
and for this combination we can arrange w1 and w2 (total 2!) between h3 and h4.

So total number of possible arrangement will be : 6 * 2 * 2 = 24

Am I correct? :roll:

bigtooth81 · Answer

H1, H2, H3 & H4 sit around the table in this order. If W1 doesn't want to sit next to H1, she has 2 choice (H2, W1, H3) and (H3,W1, H4). It's similar to W2, W3 and W4. ---> The number of possible ways = 2x2x2x2 = 16. Am I correct?

bigtooth81 · Answer

H1, H2, H3 & H4 sit around the table in this order. If W1 doesn't want to sit next to H1, she has 2 choice (H2, W1, H3) and (H3,W1, H4). It's similar to W2, W3 and W4. ---> The number of possible ways = 2x2x2x2 = 16. Am I correct?

ian7777 · Answer

Ok. I think i've got it now. Is it 12?

Of all of us, I think JPV was the closest, but he multiplied an extra 2. For each setup of husbands around the table, there are exactly 2 ways the wives can be spread around. If you map it out, you'll see that. Once one wife sits down, she dictates where everyone sits. JPV, you had 4 ways by doing 2!2!.

Since, as he showed, there are 6 ways the husbands can sit around the table, there are 12 ways all of them can sit around the table.

Isn't it great that we all start by rooting the husbands and then rotating the wives? What does that say?

By the way, I would figure out the 6 ways the husbands have a little differently: If they were sitting in a row, there would be 4! ways they could sit, but since it's a circle, and every order would have four copies (since there's no beginning or end), so divide that by 4, and get 3!=6.

mba · Answer

I think it should be 14.

4 women will take four places and 4 men will take 4 and 4 ways are there to arrange each group, so there are total 4*4 = 16 ways to arrange these people.

Now Couples can sit together in 2 ways, hence not together = 16-2 = 14

ian7777 · Answer

it's not two. I can write down on my paper here more than 2 ways.

I'm actually confident that it's 12.

I don't think the book is right in this case, boksana.

boksana · Answer

I couldn't get 2 either :roll:

The book explains the solution by picture. Sorry I cannot copy from the book. I have no a scanner. May be sometime later I'll try to draw the pic.

smandalika · Answer

Boksana
 That doesnt seem to be the correct ans. I have at least 4 combinations which satisfy the problem condition. 


Assuming the 4 hubbies H1 H2 H3 H4
and the 4 wifes are W1 W2 W3 W4

and assuming 1's , 2's 3's 4 's are a couple

4 possible cominations could be :

H1 W2 H4 W1 H3 W4 H2 W3

H4 W1 H2 W4 H3 W3 H1 W2

H4 W3 H2 W4 H1 W2 H3 W1

H4 W2 H1 W4 H3 W1 H2 W3

and the list could go on 

tell me if this is wrong???

boksana · Answer

combination 2 is impossible H3 and W3 cannot sit together

smandalika · Answer

oops sorry. but you get the picture that there could be more than 2 combinations

I've found interesting problem in a puzzle book for 6-10

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