MHB Determine how many licence plates would cost $100

  • Thread starter Thread starter WMDhamnekar
  • Start date Start date
  • Tags Tags
    Plates
AI Thread Summary
Each vehicle license plate consists of three letters followed by three digits, with costs varying based on the letters' positions in the alphabet and the digits used. The cost for digits is $n for n>0 and $10 for 0, while letters cost $1 to $26 based on their alphabetical order. The maximum cost for a license plate is $108, represented by "ZZZ 000." To find how many license plates total $100, the problem translates to distributing 8 indistinguishable items into 6 distinguishable categories, resulting in 1287 possible combinations. Thus, 1287 license plates can be produced for $100.
WMDhamnekar
MHB
Messages
376
Reaction score
28
In a certain state of a certain country, each vehicle license plates have exactly three letters followed by three digits. We are told that to produce such a license plate, it costs $n for each digit n>0 and $10 for each digit 0. For letters the cost is proportional to the position of the letter in the alphabet, namely \$1 for A, \$2for B, so on and so forth, upto $26 for Z.

Now, how to determine how many license plates would cost $100?

Answer:- I don't understand how to answer this question. I think linear programming will help here.
 
Last edited:
Mathematics news on Phys.org
This question seems to be difficult and maybe linear programming can help here.
 
Dhamnekar Winod said:
In a certain state of a certain country, each vehicle license plates have exactly three letters followed by three digits. We are told that to produce such a license plate, it costs $n for each digit n>0 and $10 for each digit 0. For letters the cost is proportional to the position of the letter in the alphabet, namely \$1 for A, \$2for B, so on and so forth, upto $26 for Z.

Now, how to determine how many license plates would cost $100?

Answer:- I don't understand how to answer this question. I think linear programming will help here.
Hello,
After working on finding out the answer to this question, eventually i suceeded. The answer to this question is 1287 license plates would cost \$100.

Justification to the answer:-

The maximum possible cost is \$108 of the license plate having ZZZ 000. Finding the number of license plates having the cost of \$100 is equivalent to finding how many ways 8 indistinguishable balls can be put into 6 distinguishable cells. So $\binom{n+r-1=13}{r=8}=1287 $ license plates.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top