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Question: How do you evaluate \(^4{P_4}\)?...

How do you evaluate 4P4^4{P_4}?

Explanation

Solution

In order to evaluate the above ,consider n=4n = 4and r=4r = 4.and use the formula of permutationsp(n,r) = nPr=n!(n−r)!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) = nCr=n!r!(n−r)!C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
p(n,r) = nPr=n!(n−r)!p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}}

Complete step by step solution:
Given4P4^4{P_4},this is of the form  nPr{\,^n}{P_r}where n=4n = 4and r=4r = 4.

To evaluate this, we will use formula of p(n,r) = nPr=n!(n−r)!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) = nPr=n!(n−r)! 4P4=4!(4−4)! 4P4=4!(0)!  p(n,r)\, = {\,^n}{P_r} = \dfrac{{n!}}{{(n - r)!}} \\\ ^4{P_4} = \dfrac{{4!}}{{(4 - 4)!}} \\\ ^4{P_4} = \dfrac{{4!}}{{(0)!}} \\\

4!4!is equivalent to 4×3×2×14 \times 3 \times 2 \times 1and we know that 0!0!is nothing but equal to 11
4P4=4×3×2×11 4P4=24  ^4{P_4} = \dfrac{{4 \times 3 \times 2 \times 1}}{1} \\\ ^4{P_4} = 24 \\\

Therefore, value of 4P4^4{P_4}is equal to 2424

Alternate:
You can alternatively find the permutation of the form nPn^n{P_n}directly by calculating n!n!.

Additional Information: 1.Factorial: The continued product of first n natural numbers is called the “n factorial “ and denoted by n!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⩽r⩽n1 \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 nPrp(n,r)\,or{\,^n}{P_r}.
p(n,r) = nPr=n!(n−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 nCrC(n,r)\,or{\,^n}{C_r}

Thus, C(n,r) or nCrC(n,r)\,or{\,^n}{C_r}= Number of ways of selecting r objects from n objects.
C(n,r) = nCr=n!r!(n−r)!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!=10! = 1.
3.Don’t forget to cross-check your answer at least once as it may contain calculation errors.