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Question: How do you evaluate \({}^{9}{{P}_{7}}\) ?...

How do you evaluate 9P7{}^{9}{{P}_{7}} ?

Explanation

Solution

The value of nPx{}^{n}{{P}_{x}} is equal to product of nCx{}^{n}{{C}_{x}} and x!x! where x!x! is the product of all the numbers from 1 to x . We know that the value of nCx{}^{n}{{C}_{x}} is equal to n!x!(nx)!\dfrac{n!}{x!\left( n-x \right)!} , so nPx{}^{n}{{P}_{x}} is equal to n!(nx)!\dfrac{n!}{\left( n-x \right)!} . By using this formula we can calculate the value of 9P7{}^{9}{{P}_{7}}

Complete step-by-step answer:
We have to evaluate 9P7{}^{9}{{P}_{7}}
We know that 9P7{}^{9}{{P}_{7}} is equal to product of nCx{}^{n}{{C}_{x}} and x!x! which is n!(nx)!\dfrac{n!}{\left( n-x \right)!}
So we can write 9P7{}^{9}{{P}_{7}} as 9!(97)!\dfrac{9!}{\left( 9-7 \right)!} which is 9!2!\dfrac{9!}{2!}
The value of 9 factorial is equal to product of all numbers from 1 to 9 and the value of 2 factorial is equal to product of all number from 1 to 2
So we can write 9!=1×2×....×99!=1\times 2\times ....\times 9 which is equal to 362880
And factorial of 2 is equal to 2
We can write 9!2!\dfrac{9!}{2!} = 3628802\dfrac{362880}{2} which is 181440

Note: nCx{}^{n}{{C}_{x}} denotes the total number of possible combination of x different object out of n different objects and nPx{}^{n}{{P}_{x}} denotes the total number of ways we can arrange x different objects out of n different objects for example the number of ways to select 4 digits out of 9 is equal to 9C4{}^{9}{{C}_{4}} and the number of ways to arrange 4 digits out of 9 is equal to 9P4{}^{9}{{P}_{4}}. In the combination we do not have to arrange for example 1, 2, 3 and 1, 3, 2 are the same in combination but they are different in permutation.