nPr Formula
The letter "P" in the nPr formula stands for "permutation" which means "arrangement". nPr formula gives the number of ways of selecting and arranging r things from the given n things. Sometimes the arrangement really matters. For example, if we have to find all the 3 digit numbers using the digits 1, 2, and 3, we would say the numbers to be 123, 132, 231, 213, 312, and 321. In this situation, the order of the digits matters to form different numbers. Let us learn the nPr formula along with a few solved examples.
What Is nPr Formula?
nPr can be written as P (n, r) (or) \(n_{ P_{r}}\) (or) \( _{n} P_{r}\). It is used to find the number of ways of selecting and arranging r different things from n different things. nPr formula is also known as permutations formula (as we call a way of choosing and arranging things to be a permutation). This formula involves factorials.
The nPr formula is:
P (n, r) (or) \(n_{ P_{r}}\) (or) \( _{n} P_{r}\) = \(\dfrac{n!}{(nr)!}\)
Solved Examples Using nPr Formula

Example 1
Find the number of 3 letter words that can be formed by rearranging the letters of the word MATH?
Solution:
The number of letters of the word MATH is n = 4.
The number of letters of each of the required words is r = 3.
Since the arrangement matters in the formation of the words, we apply the nPr formula to find the required number of 3 letter words.
P(n, r) = \(\dfrac{n!}{(nr)!}\)
P(4, 3) = \(\dfrac{4!}{(43)!}\) = \(\dfrac{4!}{1!}\) = \(\dfrac{4 \times 3 \times 2 \times 1}{1}\) = 24
Answer: The required number of 3 letter words = 24.

Example 2:
8 students have participated in a running race competition and the top three students will be awarded with the first, second, and third prizes. Find the number of ways in which the awarding can be done.
Solution:
The total number of students is n = 8.
The number of students who will be awarded = 3.
Since the arrangement among the first, second, and third prizes matters, we use the nPr formula to find the required number of ways.
P(n, r) = \(\dfrac{n!}{(nr)!}\)P(8, 3) = \(\dfrac{8!}{(83)!}\) = \(\dfrac{8!}{5!}\) = \(\dfrac{8 \times 7 \times 6 \times 5!}{5!}\) = \(8 \times 7 \times 6\) = 336
Answer: The possible number of ways = 336.