How to Use Permutations to Find Integers with Increasing Digits: 147 Example

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Homework Help Overview

The problem involves finding how many integers between 100 and 150 have three different digits arranged in increasing order, with a specific example of the number 147 provided. The context centers around the use of permutations to solve this combinatorial problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods to determine the count of valid integers, with some suggesting analytical reasoning over direct permutation application. Questions arise regarding the constraints on digit selection and the implications of increasing order.

Discussion Status

The discussion includes differing counts of valid integers, with one participant asserting a total of 32 and another confirming 18 based on their reasoning. There is no explicit consensus, but participants are engaging with the problem and exploring different interpretations of the rules.

Contextual Notes

Participants note specific rules regarding the digits, such as the requirement for increasing order and limitations on the second and third digits. There is also mention of the need for digits to be distinct and within certain ranges.

hms.tech
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Just need to learn how to use permutations in this question

How many integers between 100 and 150 have three different digits in increasing order.
One such is 147.

I solved this using a long and unnecessary method.

Can anyone guide me in how to solve it using permutations.
I got 18 as my answer.
 
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I get 32 numbers, the tens digit can contain 4 values, 0,2,3,4. While the ones digit can contain 8 values.
 
Well I don't see how you can plug this into a permutation equation and get an answer, but you can easily work it out analytically.

Rules:
1. The digits must be in increasing order.
2. The second and third digit must be greater than 1.
3. The second digit must be greater than 1 but less than 5.
4. The third digit must be greater than the second.

So that only leaves us with 2, 3, and 4 for the second digit.

When the second digit is 2 you can have 9-2 = 7 numbers in that have digits that are increasing: 23, 24, 25, 26, 27, 28, 29. When the second digit is 3 you can have 9-3 = 6 numbers: 34, 35, 36, 37, 38, 39. When the second digit is 4 you can have 9 - 4 = 5 numbers: 45, 46, 47, 48, 49. So in total you have 7 + 6 + 5 = 18 numbers. You are right.
 
12a or 13b or 14c
a>2 , b>3 , c>4
7 ways or 6 ways or 5 ways
18 ways.
 

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