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Buddy System (Posted on 2005-09-22) Difficulty: 3 of 5
Miss Honey and Miss Wormwood were taking their classes out on an outing. The two classes were of different sizes, but in each class there were the same number of girls as boys.

Miss Honey told the children in her class to form pairs, each consisting of a boy and a girl. For her own amusement, mentally calculated the number of different ways this could be done.

Miss Wormwood also told her class to form into pairs, but with at least one single-sex pair. She also insisted that the girls Matilda and Susie did not pair up together as they were the most troublesome of her girls. She too calculated the number of different ways that this could be done.

Both teachers arrived at the same answer. How many children were in each class?

No Solution Yet Submitted by Sam    
Rating: 4.3333 (6 votes)

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Solution Solution | Comment 6 of 8 |
Combinatorics...


Let H = # of students in Ms. Honey's class
Let W = # of students in Ms. Wormwood's class

The raw equation:
(H/2)! = W! / [ (W/2)! 2!^(W/2) ]  -  (W/2)!  -  (W-2)! / [ (W/2 - 1)! 2!^(W/2 - 1) ]

Reasoning:
Honey's number is simply equal to the number of permutations of a group of H/2 children.

Wormwood's number is harder to calculate.  Her number is equal to the number of permutations of a group of W children, but we must divide by 2! for each pair that is formed, and divide by (W/2)! because the order that the pairs are in doesn't matter. 

Afterwards, we subtract all the cases where there are no single-sex pairs (W/2)!.  Then, we subtract all the cases where Matilda and Susie are paired.  This number is the same as W! / [ (W/2)! 2!^(W/2) ], except plugging W-2 into W.  Note that there are no cases that are subtracted twice, since Matilda and Susie would be a single-sex pair.

Back to the equation:
(H/2)! = W! / [ (W/2)! 2!^(W/2) ]  -  (W/2)!  -  (W-2)! / [ (W/2 - 1)! 2!^(W/2 - 1) ]

Simplify the right side:
W! / [ (W/2)! 2!^(W/2) ]  -  (W/2)!  -  (W-2)! / [ (W/2 - 1)! 2!^(W/2 - 1) ]

= W! / [ (W/2)! 2^(W/2) ]  -  (W/2)!  -  W(W-2)! / [ (W/2)! 2^(W/2) ]

= W!  -  W(W-2)!   - (W/2)!
  (W/2)! 2^(W/2)

= (W/2)! * (W-1)!!  -  (W/2)! * (W-3)!!   - (W/2)!
                       (W/2)!

= (W-1)!! - (W-3)!! - (W/2)!

= (W-2)*(W-3)!! - (W/2)!

Simplest equation I could think of:
(H/2)! = (W-2)*(W-3)!! - (W/2)!

Results:
I admit it, I cheated and used excel.  But I first simplified the equation to something that I probably could have done by hand.

H is 12, W is 10, and the number each got was 720 (6!).  It is doubtful that there are other solutions (though H=W=6 would have worked, if H could equal W).

  Posted by Tristan on 2005-09-22 23:15:14
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