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Question: How do you evaluate the following expression \({}^{10}{{P}_{6}}\)?...

How do you evaluate the following expression 10P6{}^{10}{{P}_{6}}?

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 10P6{}^{10}{{P}_{6}} as
n=10;r=6n=10;r=6. We complete the multiplication and find the solution.

Complete step by step solution:
The given mathematical expression 10P6{}^{10}{{P}_{6}} 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 10P6{}^{10}{{P}_{6}}. We put the values of n=10;r=6n=10;r=6 and get
10P6=10!(10−6)!{}^{10}{{P}_{6}}=\dfrac{10!}{\left( 10-6 \right)!}.

We now solve the factorial values.

4!}{4!}=151200$$. **Therefore, the value of the combination ${}^{10}{{P}_{6}}$ is $151200$.** **Note:** There are some constraints in the form of ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$. Also, we need to remember the fact that the combination happens first even though we are finding permutation. The choosing of the $r$ objects happens first, then we arrange them. The mathematical expression is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{n!}{r!\times \left( n-r \right)!}\times r!={}^{n}{{C}_{r}}\times r!$.