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

  • Thread starter Thread starter hms.tech
  • Start date Start date
  • Tags Tags
    Permutations
AI Thread Summary
To find integers between 100 and 150 with three different digits in increasing order, the digits must follow specific rules: the second digit must be greater than 1 but less than 5, and the third digit must be greater than the second. The valid second digits are 2, 3, and 4, leading to combinations of increasing digits. For the second digit 2, there are 7 possible combinations; for 3, there are 6; and for 4, there are 5, totaling 18 valid integers. The discussion emphasizes using analytical methods over permutations to arrive at the solution.
hms.tech
Messages
246
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top