
How do you evaluate $^4{P_4}$?
Answer
558.9k+ views
Hint:In order to evaluate the above ,consider $n = 4$and $r = 4$.and use the formula of permutations$p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$to find the number of permutations that can be made by taking 4 things from the all number of things which is also 4.the answer you will get will have all possible arrangements possible included.
Formula:
$C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Complete step by step solution:
Given$^4{P_4}$,this is of the form ${\,^n}{P_r}$where $n = 4$and $r = 4$.
To evaluate this, we will use formula of $p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$
So, Putting the value of n and r in the above formula
$
p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}} \\
^4{P_4} = \dfrac{{4!}}{{(4 - 4)!}} \\
^4{P_4} = \dfrac{{4!}}{{(0)!}} \\
$
$4!$is equivalent to \[4 \times 3 \times 2 \times 1\]and we know that $0!$is nothing but equal to $1$
$
^4{P_4} = \dfrac{{4 \times 3 \times 2 \times 1}}{1} \\
^4{P_4} = 24 \\
$
Therefore, value of $^4{P_4}$is equal to $24$
Alternate:
You can alternatively find the permutation of the form $^n{P_n}$directly by calculating $n!$.
Additional Information:1.Factorial: The continued product of first n natural numbers is called the “n factorial “ and denoted by $n!$.
2.Permutation: Each of the arrangements which can be made by taking some or all of a number of things are called permutations.
If n and r are positive integers such that $1 \leqslant r \leqslant n$, then the number of all
permutations of n distinct or different things, taken r at one time is denoted by the symbol
$p(n,r)\,or{\,^n}{P_r}$.
$p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$
3.Combinations: Each of the different selections made by taking some or all of a number of objects irrespective of their arrangement is called a combination.
The combinations number of n objects, taken r at one time is generally denoted by
$C(n,r)\,or{\,^n}{C_r}$
Thus, $C(n,r)\,or{\,^n}{C_r}$= Number of ways of selecting r objects from n objects.
$C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Note: 1. Factorials of proper fractions or negative integers are not defined. Factorial n defined only for whole numbers.
2.Meaning of Zero factorial is senseless to define it as the product of integers from 1 to zero. So, we
define it as $0! = 1$.
3.Don’t forget to cross-check your answer at least once as it may contain calculation errors.
Formula:
$C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Complete step by step solution:
Given$^4{P_4}$,this is of the form ${\,^n}{P_r}$where $n = 4$and $r = 4$.
To evaluate this, we will use formula of $p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$
So, Putting the value of n and r in the above formula
$
p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}} \\
^4{P_4} = \dfrac{{4!}}{{(4 - 4)!}} \\
^4{P_4} = \dfrac{{4!}}{{(0)!}} \\
$
$4!$is equivalent to \[4 \times 3 \times 2 \times 1\]and we know that $0!$is nothing but equal to $1$
$
^4{P_4} = \dfrac{{4 \times 3 \times 2 \times 1}}{1} \\
^4{P_4} = 24 \\
$
Therefore, value of $^4{P_4}$is equal to $24$
Alternate:
You can alternatively find the permutation of the form $^n{P_n}$directly by calculating $n!$.
Additional Information:1.Factorial: The continued product of first n natural numbers is called the “n factorial “ and denoted by $n!$.
2.Permutation: Each of the arrangements which can be made by taking some or all of a number of things are called permutations.
If n and r are positive integers such that $1 \leqslant r \leqslant n$, then the number of all
permutations of n distinct or different things, taken r at one time is denoted by the symbol
$p(n,r)\,or{\,^n}{P_r}$.
$p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}$
3.Combinations: Each of the different selections made by taking some or all of a number of objects irrespective of their arrangement is called a combination.
The combinations number of n objects, taken r at one time is generally denoted by
$C(n,r)\,or{\,^n}{C_r}$
Thus, $C(n,r)\,or{\,^n}{C_r}$= Number of ways of selecting r objects from n objects.
$C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Note: 1. Factorials of proper fractions or negative integers are not defined. Factorial n defined only for whole numbers.
2.Meaning of Zero factorial is senseless to define it as the product of integers from 1 to zero. So, we
define it as $0! = 1$.
3.Don’t forget to cross-check your answer at least once as it may contain calculation errors.
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