Math Problem Statement

Consider the letters of the word REPRESENTATIVE. The letters that appear more than once in this word are treated as being identical. 15.2.1 How many word arrangements can be made with the letters of this word? 15.2.2 How many word arrangements can be made if the letter E must be in the first position? 15.2.3 How many word arrangements can be made if the letter E must NOT be in the first position?

Solution

Let's break down and solve the questions one by one, based on the word "REPRESENTATIVE."

Step 1: Analyze the word "REPRESENTATIVE"

The word "REPRESENTATIVE" has 12 letters. The frequency of the repeated letters is as follows:

  • E appears 4 times.
  • R appears 2 times.
  • T appears 2 times.

The rest of the letters (P, S, N, A, I, V) appear only once.

15.2.1: How many word arrangements can be made with the letters of this word?

To find the total number of distinct arrangements of the letters, we use the formula for permutations of multiset: Number of arrangements=12!4!×2!×2!\text{Number of arrangements} = \frac{12!}{4! \times 2! \times 2!} Where:

  • 12!12! is the factorial of the total number of letters.
  • 4!4! accounts for the 4 repeated "E"s.
  • 2!2! accounts for the 2 repeated "R"s.
  • 2!2! accounts for the 2 repeated "T"s.

Calculating this: 12!=479,001,600,4!=24,2!=212! = 479,001,600, \quad 4! = 24, \quad 2! = 2 Number of arrangements=479,001,60024×2×2=479,001,60096=4,989,600\text{Number of arrangements} = \frac{479,001,600}{24 \times 2 \times 2} = \frac{479,001,600}{96} = 4,989,600

So, the total number of distinct word arrangements is 4,989,600.

15.2.2: How many word arrangements can be made if the letter E must be in the first position?

If one of the E's is fixed in the first position, we are left with 11 remaining letters, which include 3 E's, 2 R's, and 2 T's. So, we apply the permutation formula for the remaining letters: Number of arrangements=11!3!×2!×2!\text{Number of arrangements} = \frac{11!}{3! \times 2! \times 2!} Where:

  • 11!=39,916,80011! = 39,916,800
  • 3!=63! = 6
  • 2!=22! = 2

Number of arrangements=39,916,8006×2×2=39,916,80024=1,663,200\text{Number of arrangements} = \frac{39,916,800}{6 \times 2 \times 2} = \frac{39,916,800}{24} = 1,663,200

Thus, the number of word arrangements with E in the first position is 1,663,200.

15.2.3: How many word arrangements can be made if the letter E must NOT be in the first position?

To solve this, we subtract the number of arrangements where E is in the first position (from 15.2.2) from the total number of arrangements (from 15.2.1): Arrangements without E in first position=4,989,6001,663,200=3,326,400\text{Arrangements without E in first position} = 4,989,600 - 1,663,200 = 3,326,400

Therefore, the number of word arrangements where E is not in the first position is 3,326,400.


Would you like more detailed explanations, or do you have any other questions?

Here are 5 related questions for further practice:

  1. How many arrangements can be made if both R's must be adjacent?
  2. How many arrangements can be made if the word starts and ends with a vowel?
  3. How many distinct arrangements are possible if only one of the E's can be used?
  4. How many arrangements can be made if the letter T must always be in the middle position?
  5. How many arrangements can be made if all vowels must appear together?

Tip: When dealing with repeated elements in permutations, always remember to divide by the factorial of the number of repeated items to avoid overcounting.

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Math Problem Analysis

Mathematical Concepts

Permutations of Multisets
Factorial

Formulas

Number of arrangements = n! / (k1! × k2! × ... × kr!)
Where n is the total number of elements, and ki is the factorial of the number of times a repeated element occurs

Theorems

Permutation Theorem for Multisets

Suitable Grade Level

Grade 11-12