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[tex]__________________________[/tex]
[tex]\sf _n P_r = \dfrac{n!}{(n-r)!}[/tex]
where,
[tex]\sf _{10} P_3 = \dfrac{10!}{(10-3)!}[/tex]
[tex]\sf _{10} P_3 = \dfrac{10!}{7!}[/tex]
[tex]\sf _{10} P_3 = \dfrac{3628800}{5040}[/tex]
[tex]\sf _{10} P_3 = 720[/tex]
Answer:
[tex]\sf _{10} P_3 = 720[/tex]
∴ There are 720 ways that the 1st, 2nd and 3rd places can be decided.
Answer:
To determine in how many different ways the first, second, and third places can be decided in a horse race with ten horses, we can use the concept of permutations.
The number of ways in which we can select the first, second, and third places out of ten horses is given by the permutation formula for selecting r objects out of n distinct objects, which is:
P(n, r) = n! / (n - r)!
In this case, we want to select 3 horses from 10 for the first, second, and third places respectively. So, applying the formula:
P(10, 3) = 10! / (10 - 3)!
= 10! / 7!
Calculating the factorial values:
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 3628800
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040
Now, substitute these values back:
P(10, 3) = 3628800 / 5040
= 720
Therefore, the number of different ways in which the first, second, and third places can be decided in a race with ten horses is 720 ways.
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