How many ways do 10 people come in 1st, 2nd and 3rd place in race once around the track?
Answer
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Hint: The permutation of n things taken r at a time is given by ${}^{n}{{P}_{r}}$ . The given question asks to find out the permutations of 10 things taken 3 at a time. We use the formula of ${}^{n}{{P}_{r}}$ that is $\dfrac{n!}{\left( n-r \right)!}$ where n =10 and r=3 as per the question to get the required result.
Complete step by step solution:
We need to find the number of ways 10 people can come in 1st, 2nd, and 3rd place in a race once around the track. Here, in our case order is important so we will be solving the given question using permutations.
Permutations usually are the number of ways the objects can be ordered or selected.
Order is usually a constraint in permutations, unlike combinations.
The permutation of n things or items taken r at a time is denoted by ${}^{n}{{P}_{r}}$
n = no of things
r = no of things taken at the time.
The combination of n things taken r at a time or ${}^{n}{{P}_{r}}$ is given by the formula
$\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
The exclamation mark in the above formula denotes factorial.
Factorial of a number in mathematics is the product of all the positive numbers less than or equal to a number.
The multiplication happens to a given number down to the number one or till the number one is reached.
Example: Factorial of n is n! and the value of n! is n! = $n\times (n-1)\times (n-2)\times ………\times 1$.
In the given question,
$n=10$ ;
$r=3$
Substituting the same in the formula, we get,
$= \dfrac{10!}{\left( 10-3 \right)!}$
$= \dfrac{10!}{7!}$
Writing the values of the factorials, we get,
$= \dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 6\times 5\times 4\times 3\times 2\times 1}$
Cancelling the common terms on numerator and denominator, we get,
$= 10\times 9\times 8$
Simplifying the above expression, we get,
$= 720$
$\therefore$ There are 720 possible ways for 10 people to come 1st, 2nd and 3rd place in a race once around the track.
Note: The factorial of a number n should only go down to 1 and not zero. The order of selection or sequence matters in the case of permutations, unlike combinations. Common mistakes like double counting, confusion of values should be avoided to get precise results.
Complete step by step solution:
We need to find the number of ways 10 people can come in 1st, 2nd, and 3rd place in a race once around the track. Here, in our case order is important so we will be solving the given question using permutations.
Permutations usually are the number of ways the objects can be ordered or selected.
Order is usually a constraint in permutations, unlike combinations.
The permutation of n things or items taken r at a time is denoted by ${}^{n}{{P}_{r}}$
n = no of things
r = no of things taken at the time.
The combination of n things taken r at a time or ${}^{n}{{P}_{r}}$ is given by the formula
$\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
The exclamation mark in the above formula denotes factorial.
Factorial of a number in mathematics is the product of all the positive numbers less than or equal to a number.
The multiplication happens to a given number down to the number one or till the number one is reached.
Example: Factorial of n is n! and the value of n! is n! = $n\times (n-1)\times (n-2)\times ………\times 1$.
In the given question,
$n=10$ ;
$r=3$
Substituting the same in the formula, we get,
$= \dfrac{10!}{\left( 10-3 \right)!}$
$= \dfrac{10!}{7!}$
Writing the values of the factorials, we get,
$= \dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 6\times 5\times 4\times 3\times 2\times 1}$
Cancelling the common terms on numerator and denominator, we get,
$= 10\times 9\times 8$
Simplifying the above expression, we get,
$= 720$
$\therefore$ There are 720 possible ways for 10 people to come 1st, 2nd and 3rd place in a race once around the track.
Note: The factorial of a number n should only go down to 1 and not zero. The order of selection or sequence matters in the case of permutations, unlike combinations. Common mistakes like double counting, confusion of values should be avoided to get precise results.
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