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Question: How do you evaluate \({}^{3}{{P}_{1}}\)?...

How do you evaluate 3P1{}^{3}{{P}_{1}}?

Explanation

Solution

We first discuss the general form of permutation and its general meaning with the help of variables. We express the mathematical notion with respect to the factorial form of nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. Then we place the values for 3P1{}^{3}{{P}_{1}} as n=3;r=1n=3;r=1. We complete the multiplication and find the solution.

Complete step-by-step solution:
The given mathematical expression 3P1{}^{3}{{P}_{1}} is an example of permutation.
We first try to find the general form of permutation and its general meaning and then we put the values to find the solution.
The general form of permutation is nPr{}^{n}{{P}_{r}}. It’s used to express the notion of choosing rr objects out of nn objects and then arranging those rr objects. The value of nPr{}^{n}{{P}_{r}} expresses the number of ways the permutation of those objects can be done.
The simplified form of the mathematical expression nPr{}^{n}{{P}_{r}} is nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}.
Here the term n!n! defines the notion of multiplication of first n natural numbers.
This means n!=1×2×3×....×nn!=1\times 2\times 3\times ....\times n.
The arrangement of those chosen objects is not considered in case of combination. That part is involved in permutation.
Now we try to find the value of 3P1{}^{3}{{P}_{1}}. We put the values of n=3;r=1n=3;r=1 and get 3P1=3!(3−1)!{}^{3}{{P}_{1}}=\dfrac{3!}{\left( 3-1 \right)!}.
We now solve the factorial values 3P1=3!2!=3×2×12×1=3{}^{3}{{P}_{1}}=\dfrac{3!}{2!}=\dfrac{3\times 2\times 1}{2\times 1}=3.
Therefore, the value of the combination 3P1{}^{3}{{P}_{1}} is 3.

Note: There are some constraints in the form of nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. The general conditions are n≥r≥0;n≠0n\ge r\ge 0;n\ne 0. Also, we need to remember the fact that the combination happens first even though we are finding permutation. The choosing of the rr objects happens first, then we arrange them. The mathematical expression is nPr=n!(n−r)!=n!r!×(n−r)!×r!=nCr×r!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{n!}{r!\times \left( n-r \right)!}\times r!={}^{n}{{C}_{r}}\times r!.