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 Buddy System (Posted on 2005-09-22)
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 Set-up Comment 8 of 8 |

2H children in Honey's class, 2W in Wormwood's.

CH = Number of ways for arranging Honey's class = H! (as in duh!)

Wormwood's problem is more interesting:

The number of ways of selecting a pair from among W - 2N boys is

(W-2N) (W-2N-1) / 2

Hence the number of ways to form N pairs of boys is

BN = W! / [(W-2N)! 2^N].

The number of ways to form N pairs of girls, each of which includes the forbidden combination (Mathilde, Susie) is

FN = (W-2)! / [(W-2N)! 2^(N-1)]

this is calculated by taking the number of ways of forming N-1 pairs from among W-2 girls (the pair D,S already having been assigned).

Therefore the number of allowed combinations of N pairs of girls is

GN = BN - FN

So, the number of ways of arranging Wormwood's class so that there are 2N same sex pairs and W-2N mixed pairs is

QN = BN * GN * (W-2N)!

The Wormwood number is thus

T = Sum QN    where the sum runs from N=1 to N = [W/2]; and, according the the premise of the problem, T equals a factorial of some number H.

... more to come..

Edited on March 28, 2008, 8:12 am
 Posted by FrankM on 2008-03-28 08:10:46

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