Chapter 9, Problem 18RE

Solution

i.

To state: The number of distinguishable that can be made from the word FLORIDA.

The resultant answer is 5040.

Given information:

The given word is “FLORIDA” consisting seven letters.

Explanation:

There are total 7 letters in the word “FLORIDA”. The value of $n=7$ .

As no letter is repeated in the word FLORIDA, then the number of distinguishable permutation will become:

  $\begin{array}{c}n!=7!\\ =7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\\ =5040\end{array}$

Then total numbers of distinguishable permutations are 5040.

ii.

To calculate: The number of distinguishable that can be made from the word TALLAHASSEE.

The resultant answer is $831,600$ .

Given information:

The given word is “TALLAHASSEE” consisting eleven letters.

Formula used: If an n -set contains $n_1$ objects of first kind, $n_2$ objects of a second kind, and so on, with $n_1+n_2+\cdots +n_k=n$ , then the number of distinguishable permutations of the n -set is given by $\frac{n!}{n_1!n_2!\cdots n_k!}$ .

Calculation:

There are total 11 letters in the word “TALLAHASSEE”. There are 3 A’s, 2 L’s, 2 E’s, and 2 S’s in this word.

The values are $n=11,n_1=3,n_2=2,n_3=2,\text{and}{n}_4=2$ ,

Use the formula $\frac{n!}{n_1!n_2!\cdots n_k!}$ and substitute the values,

  $\frac{n!}{n_1!n_2!\cdots n_k!}=\frac{11!}{3!2!2!2!}$

Now expand the factorial,

  $\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3!}{3!\cdot 2\cdot 2\cdot 2}=831,600$

Then total numbers of distinguishable permutations are $831,600$ .