In a dog show, there will be a gold and a silver medal awarded. There are
4
finalists. In how many different ways can the
2
medals be awarded?
ab
ac
ad
ba
bc
bd
ca
cb
cd
da
db
dc
and all that equals 4!/2!
To find the number of different ways the medals can be awarded, we use the concept of permutations.
Since there are 4 finalists and 2 medals to be awarded, we need to find the number of permutations of 4 items taken 2 at a time.
This can be calculated using the formula for permutations:
P(n, r) = n! / (n - r)!
where n is the total number of items and r is the number of items taken at a time.
In this case, n = 4 and r = 2.
P(4, 2) = 4! / (4 - 2)!
= 4! / 2!
= (4 * 3 * 2 * 1) / (2 * 1)
= 24 / 2
= 12
Therefore, there are 12 different ways the gold and silver medals can be awarded in the dog show.
To find the number of different ways the gold and silver medals can be awarded, we can use the concept of combinations.
Since there are 4 finalists and 2 medals to be awarded, we need to select 2 finalists for the medals. The order in which the finalists are selected does not matter, as both medals have the same value.
The number of ways to select 2 finalists out of 4 can be calculated using the formula for combinations, which is given by:
C(n, k) = n! / (k! * (n-k)!)
where n is the total number of items and k is the number of items being selected.
In this case, n = 4 (the number of finalists) and k = 2 (the number of medals). Plugging these values into the formula, we get:
C(4, 2) = 4! / (2! * (4-2)!)
= 4! / (2! * 2!)
= (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1))
= 24 / 4
= 6
Therefore, there are 6 different ways the gold and silver medals can be awarded in the dog show.