Algebra 2 - Permutations

Introduction - Fundamental Counting Principle and Permutation

  • The fundamental counting principle states that if one thing can occur in m ways, a second occurs in n ways, and a third thing can occur in r ways, then the sequence of things can occur in m·n·r ways.
  • Factorial Notation defines that the product of the positive integers from 1 to n inclusive occurs typically in the field of mathematics and is denoted by the notation n!, which is termed orally as "n factorial."
  • It is a given fact that 0! = 1.
  • n!=nn-1n-2n-3...321
  • A Permutation is an arrangement of n objects in a given order. An arrangement of any rn of these objects in a given order is a permutation of n taken r at a time.
  • Generally, the formula of Permutation is given by Prn=n!n-r! or Pn, r=n!n-r!.


Some Solved Examples involving the Fundamental Counting Principle:

Example 1. How many three-digit numbers can be formed using the digits 0, 4, 7, and 8 if no digit is repeated?

Solution: We can count the total possible permutations of three-digit numbers by using the blanks below that describe the position of the unit digit, the tens digit, and the hundreds digit.

The hundredth digit can be done in 3 ways since 0 can't be the first digit. The tenth digit can be done in 3 ways. The units digit can be done in 2 ways.

Using the fundamental counting principle, we have: n1·n2·n3=3·3·2=18 three-digit numbers


Example 2. How many four-digit even numbers greater than 3,000 can be formed using only the digits 1, 2, 3, and 4 if repetition is allowed?


Use the concept of the Fundamental Counting Principle.

For four-digit numbers, we have: 2·4·4·2=64

For three-digit numbers, we have: 4·4·2=32

For two-digit numbers, we have: 4·2=8

One-digit numbers can have two numbers only.

Thus, the total number of four-digit even numbers formed greater than 3,000 is computed as: 64+32+8+2=106.

Solved Examples

Consider the letters x, y, and z. This means all their possible arrangements are x,y,z, x,z,y, z,y,x, z,x,y,y,z,x, y,x,z. This implies permutation 3 taken 3, which is 6.

Example 1. Evaluate P47?

Using the formula Prn=n!n-r!, we have n=7 and r=4



Example 2. Find the number of permutations of 7 letters, a, b, c, d, e, f, and g taken three at a time. 

Using the Fundamental Counting Principle, we have: n1·n2·n3=7·6·5=210

Using Permutation, we have: P7,3=7!7-3!=210


Example 3. In how many ways can the letters in the word ENGINEERING be arranged?

Pn,nr=11!2!2!3!3!=277,200 permutations

Letters E and N are repeated three times, and letters I and N are repeated twice.


Example 4. How many ways can one select three cards in succession from a deck of cards with replacements? Without replacement?

With Replacement: 52·52·52 or 523=140,608 ways

Without Replacement: 52·51·50=132,600 ways


Example 5. Suppose nine mathematics and six physics books will be arranged on a shelf. How many ways can such books be arranged if the books of the same subject shall be placed side by side?



Example 6. How many ways can 7sevenmen be seated at a circular table such that two men must always sit beside each other?

n-1!·Pn,r=7-1!·P2,2=1440 ways

Cheat Sheet

  • The formula P=nr is used when order is important to compute permutations and repetition is allowed. 
  • The formula Prn=n!n-r! is used when order matters and repetition is not allowed.

Blunder Areas

  • The number of permutations is larger than the number of combinations.
  • The concept of permutations stresses that order matters, while the concept of combinations uses no order of arrangement. 
  • 0!0. It is a basic idea that 0!=1.
  • In the formula Prn=n!n-r!, rn.
  • In solving problems involving permutations and combinations, always be careful in analyzing the type of situation involved, whether it involves an order of arrangement or not.