Answer
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Hint: This question is based on the permutation. The different arrangements which can be made out of a given number of objects by taking some or all at a time, are called permutation. And for the calculation we use a formula for permutation of ‘n’ objects for ‘r’ selection of objects is given by:
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Where $!$ is the notation for the factorial.
So to solve the given question we first use this formula and after further simplifying we get the required answer.
Complete step by step solution:
Given ${}^{10}{P_r} = 720$
In this question we have to compute the value of ‘r’.
Now for solving this first we use the formula of permutation i.e.
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ On the left side of the question.
Here we put $n = 10$
$\dfrac{{10!}}{{\left( {10 - r} \right)!}} = 720$
Now we solve $10!$ on left side and taking factor of \[720\] on right side
$\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {10 - r} \right)!}}$$ = 10 \times 9 \times 8$
$ \Rightarrow \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{10 \times 9 \times 8}} = \left( {10 - r} \right)!$
Now we simplify left side by cancelling the similar term
$ \Rightarrow 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \left( {10 - r} \right)!$
Here we can also write left side as $7!$
$ \Rightarrow 7! = \left( {10 - r} \right)!$
On comparing the both sides
$ \Rightarrow 7 = 10 - r$
Now we shift ‘r’ on left side and $7$ on right side
$ \Rightarrow r = 10 - 7$
$ \Rightarrow r = 3$
Hence, if ${}^{10}{P_r} = 720$ the required value of ‘r’ is $3$.
Note:
In this type of question we should know the formula of permutation. And in the formula of permutation factorial notation $\left( {!or\left| \!{\underline {\,
{} \,}} \right. } \right)$ has been used. If n be a positive integer. Then the continued product of first ‘n’ natural numbers is called factorial n. It is denoted by $\left| \!{\underline {\,
n \,}} \right. $or $n!$. Here we should remember $\left| \!{\underline {\,
0 \,}} \right. = 1$. The question of permutations is usually based on the arrangement. So many different types of questions can be made on the basis of permutation. And for solving such questions we will have to use the formula of permutation.
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Where $!$ is the notation for the factorial.
So to solve the given question we first use this formula and after further simplifying we get the required answer.
Complete step by step solution:
Given ${}^{10}{P_r} = 720$
In this question we have to compute the value of ‘r’.
Now for solving this first we use the formula of permutation i.e.
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ On the left side of the question.
Here we put $n = 10$
$\dfrac{{10!}}{{\left( {10 - r} \right)!}} = 720$
Now we solve $10!$ on left side and taking factor of \[720\] on right side
$\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {10 - r} \right)!}}$$ = 10 \times 9 \times 8$
$ \Rightarrow \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{10 \times 9 \times 8}} = \left( {10 - r} \right)!$
Now we simplify left side by cancelling the similar term
$ \Rightarrow 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \left( {10 - r} \right)!$
Here we can also write left side as $7!$
$ \Rightarrow 7! = \left( {10 - r} \right)!$
On comparing the both sides
$ \Rightarrow 7 = 10 - r$
Now we shift ‘r’ on left side and $7$ on right side
$ \Rightarrow r = 10 - 7$
$ \Rightarrow r = 3$
Hence, if ${}^{10}{P_r} = 720$ the required value of ‘r’ is $3$.
Note:
In this type of question we should know the formula of permutation. And in the formula of permutation factorial notation $\left( {!or\left| \!{\underline {\,
{} \,}} \right. } \right)$ has been used. If n be a positive integer. Then the continued product of first ‘n’ natural numbers is called factorial n. It is denoted by $\left| \!{\underline {\,
n \,}} \right. $or $n!$. Here we should remember $\left| \!{\underline {\,
0 \,}} \right. = 1$. The question of permutations is usually based on the arrangement. So many different types of questions can be made on the basis of permutation. And for solving such questions we will have to use the formula of permutation.
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