## 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 $m\xb7n\xb7r$ ways.- Factorial Notation defines that the product of the positive integers from $1ton$ inclusive occurs typically in the field of mathematics and is denoted by the notation $\mathit{n}\mathbf{!}$, which is termed orally as $"nfactorial."$
- It is a given fact that $0!$ = 1.
- $n!=n\left(n-1\right)\left(n-2\right)\left(n-3\right)...\left(3\right)\left(2\right)\left(1\right)$
- A Permutation is an arrangement of $n$ objects in a given order. An arrangement of any $r\le n$ 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 ${}_{n}P_{r}=\frac{n!}{\left(n-r\right)!}$ or $P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$.

**Some Solved Examples involving the Fundamental Counting Principle:**

**Example 1.** How many three-digit numbers can be formed using the digits $0,4,7,and8$ 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: ${n}_{1}\xb7{n}_{2}\xb7{n}_{3}=3\xb73\xb72=\mathbf{18}\mathbf{}\mathit{t}\mathit{h}\mathit{r}\mathit{e}\mathit{e}\mathbf{-}\mathit{d}\mathit{i}\mathit{g}\mathit{i}\mathit{t}\mathbf{}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathit{s}$

**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?

Solution:

Use the concept of the Fundamental Counting Principle.

For four-digit numbers, we have: $2\xb74\xb74\xb72=64$

For three-digit numbers, we have: $4\xb74\xb72=32$

For two-digit numbers, we have: $4\xb72=8$

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 $x,y,andz.$This means all their possible arrangements are $\left\{x,y,z\right\},\left\{x,z,y\right\},\left\{z,y,x\right\},\left\{z,x,y\right\},\left\{y,z,x\right\},\left\{y,x,z\right\}$. This implies permutation 3 taken 3, which is 6.

**Example 1.** Evaluate ${}_{7}P_{4}$?

Using the formula ${}_{n}P_{r}=\frac{n!}{\left(n-r\right)!}$, we have $n=7andr=4$.

${}_{7}P_{4}=\frac{7!}{\left(7-4\right)!}=\frac{7\xb76\xb75\xb74\xb73\xb72\xb71}{3\xb72\xb71}=840$

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

Using the Fundamental Counting Principle, we have: ${n}_{1}\xb7{n}_{2}\xb7{n}_{3}=7\xb76\xb75=210$

Using Permutation, we have: $P\left(7,3\right)=\frac{7!}{\left(7-3\right)!}=210$

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

$P\left(n,{n}_{r}\right)=\frac{11!}{2!2!3!3!}=277,200permutations$

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: $52\xb752\xb752or{52}^{3}=140,608ways$

Without Replacement: $52\xb751\xb750=132,600ways$

**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?

$P(9,9)\xb7P\left(6,6\right)\xb72!=522,547,200$

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

$\left(n-1\right)!\xb7P\left(n,r\right)=\left(7-1\right)!\xb7P\left(2,2\right)=1440ways$

## Cheat Sheet

- The formula $P={n}^{r}$ is used when order is important to compute permutations and repetition is allowed.
- The formula ${}_{n}P_{r}=\frac{n!}{\left(n-r\right)!}$ 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.
- $0!\ne 0$. It is a basic idea that $0!=1$.
- In the formula ${}_{n}P_{r}=\frac{n!}{\left(n-r\right)!},$$r\le n$.
- 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.

- Keith Madrilejos
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