Loading the starting gate for a horse race
A,B,C,D,E,F,G,H, and I are 9 horses in a race. Three of them are two-year-olds, three are three-year-olds and three are four-year-olds.The starting gate has stalls numbered from 1 through 9 and the horses are to enter the stalls one by one such that a horse can choose any vacant stall and such that at no time does the number of three-year-olds in the starting gate exceed the number of two-year-olds nor does the number of four-year-olds exceed the number of three-year-olds in the gate . How many ways can this be done?
The answer is 3292047360.
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1 solution
We note that permuting the order in which the two years olds entered, the order in which three-year-olds entered and the order in which the four-year-olds entered in any given admissible entry sequence gives rise
to 3! times 3! times 3! = 216 admissible entry sequences ( including the original) and permuting the stall numbers give rise to 9! admissible entry sequences ( including the original). Now 216 times 9!=78382080
Let the entry number of a horse denote when it entered a stall e.g., if a horse was the 4th horse to enter his entry number is 4.Now every 3 by 3 Standard Young Tableau corresponds to a unique admissible entry sequence since it contains the numbers 1 through 9 in which the row and column entries increase so that the first row can be viewed from left to right as the entry numbers of the two-year-olds, the second row can be viewed from left to right as the entry numbers of the three-year-olds and the third row similarly as the entry numbers of the four-year-olds.( The first column can thus be viewed as the entry numbers of the first two year old, the first three year old and the first four year old with the second column corresponding to the entry numbers of the second two year old, the second three year old and the second four year old and the third column as the entry numbers of the third two year old, the third three year old and the third four year old (Clearly the first two-year-old enters before the second two-year-old who in turn enters before the third two-year-old and the first two-year-old enters before the first three- year- old and so on .). Now the hooklengths of this tableau are 1,2,3,2,3,4,3,4,5 and their product is 8640 and so by the hooklength formula the number of such standard tableaux is 9! divided by 8640 which gives 42
The answer is thus 42 times 78382080 = 3292047360