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## Algebra 1 word problem about horses.

At a horse show, ribbons are given for first, second,third, and fourth places. There are 20 horses in the show. How many different arrangements of four winning horses are possible?

I agree with the technical solution from Priti, and I would like to add to how to reason such a problem.

Let's say that we have a total of 5 horses instead of 20. We will reason the problem in this simplified form and then the reasoning is the same for the larger problem. Let's call the horses: John, Mary, Peter, Jack, and Sara. They compete at the show and only 4 of them can win the ribbons for 1st, 2nd, 3rd, and 4th. In how many ways can the race end? (in other words, how many arrangements of 4 different horses are possible in this case?).

Notice that 2 different horses cannot win the same prize at the same time. If Peter comes 1st, he cannot be 2nd at the same time. Let's say that Mary came 2nd, Sara 3rd, and Jack 4th (John did not get anything).

So the arrangement is in this case: Peter(1), Mary(2), Sara(3), Jack(4).

Notice that if the winners were Mary(1), Peter(2), Sara(3), Jack(4), then the arrangement would have been different, so order matters.

Also notice that John was not included. So different arrangements are possible depending on who the winners are, there are only 4 places, order matters, and two horses cannot win the same prize at the same time. Hence- to count the number of arrangements, we need to use the formula for permutations.

For a total of 5 horses: 5 P 4 = using factorials 5 !/(5-4)! = 5*4*3*2=120

For a total of 20 (replace 5 with a larger number of competitor horses): 20 P 4, as calculated in Priti's solution.

Need to do Permutation.

nPr    ---->  20 P 4  = 51680

or

20 * 19  * 18 * 17  =   51680