Question

Evaluate the following expression $\dfrac{n!}{(n-r)!}$ when n = 9 and r = 5.

Answer 1

Hint: In this question, we use the formula to find the factorial of a number is given by $n!=n\times (n-1)\times (n-2)\times .......\times 3\times 2\times 1$. The factorial is the product of all integers less than or equal to n but greater than or equal to 1.

Complete step by step answer:
In Mathematics, factorial is a simple thing. Factorials are just products. An exclamation mark indicates the factorial. Factorial is a multiplication operation of natural numbers with all the natural numbers that are less than it.
The multiplication of all positive integers says n, that will be smaller than or equivalent to n is known as the factorial. The factorial of a positive integer is represented by the symbol $n!$ .
From the formula of factorial, the recurrence relation for the factorial of a number is defined as the product of factorial number and factorial of that number minus 1. It is given by
$n!=n\times (n-1)!$
Let us consider the given expression $\dfrac{n!}{(n-r)!}$ and put the given values n = 9 and r = 5, then we have
$\dfrac{n!}{(n-r)!}=\dfrac{9!}{(9-5)!}=\dfrac{9!}{4!}.....................(1)$
By definition of the factorial,
$9!=9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$ and $4!=4\times 3\times 2\times 1$
Equation (1) becomes
$\dfrac{n!}{(n-r)!}=\dfrac{9!}{(9-5)!}=\dfrac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}$
Cancelling some terms, we get
$\dfrac{n!}{(n-r)!}=\dfrac{9!}{(9-5)!}=9\times 8\times 7\times 6\times 5$
$\dfrac{n!}{(n-r)!}=\dfrac{9!}{(9-5)!}=15120$
Hence, the value of the expression $\dfrac{n!}{(n-r)!}$ when n =9 and r =5 is 15120.

Note: The factorial of 0 is 1. According to the convention of empty product, the result of multiplying no factors is a nullary product. It means that the convention is equal to the multiplicative identity.