Be careful with your licence plates!
In Kansas, vehicle licence plates each have exactly three letters followed by three digits. We are told that to produce such a licence plate, it costs $ n for each digit n > 0 and $ 1 0 for each digit 0 . For letters, the cost is proportional to the position of the letter in the alphabet, namely, $ 1 for A , $ 2 for B and so on, up to $ 2 6 for Z .
Determine how many different licence plates would cost exactly $ 1 0 0 .
The answer is 1287.
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4 solutions
Could you explain why it is the same?
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The indistinguishable "balls" in this case are decrementations of each character by 1. (8 decrementations means that the license plate is guaranteed to be $100) There is allowed to be any number of them in a character (putting all of them in a digit is only 2, all in a letter is R), there are six characters, and the order matters. So, 8 identical balls in 6 different boxes.
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We can also use the tables below to find plates for Digit parts and then as above.
The answer is the coefficient of x 1 0 0 in the following polynomial: ( x + x 2 + . . . + x 1 0 ) 3 ( x + x 2 + . . . + x 2 6 ) 3 Which is by factorization the coefficient of x 9 4 in the following polynomial: ( 1 + x + . . . + x 9 ) 3 ( 1 + x + . . . + x 2 5 ) 3 Which is the coefficient of x 8 in the same polynomial since this polynomial is Palindromic : ( 1 + x + . . . + x 9 ) 3 ( 1 + x + . . . + x 2 5 ) 3 Which is by removing power more than 8 the coefficient of x 8 in the following polynomial: ( 1 + x + . . . + x 8 ) 6 ( 1 + x + . . . + x 8 ) 6 = ( 1 − x ) 6 ( 1 − x 9 ) 6 let's take a detour through analysis: ( 1 − x 9 ) 6 = 1 + O ( x 9 ) and by Taylor series: ( 1 − x ) 6 1 = 1 + 6 x + 2 1 x 2 + 5 6 x 3 + 1 2 6 x 4 + 2 5 2 x 5 + 4 6 2 x 6 + 7 9 2 x 7 + 1 2 8 7 x 8 + O ( x 9 ) so that : ( 1 + x + . . . + x 8 ) 6 = 1 + 6 x + 2 1 x 2 + 5 6 x 3 + 1 2 6 x 4 + 2 5 2 x 5 + 4 6 2 x 6 + 7 9 2 x 7 + 1 2 8 7 x 8 + O ( x 9 )
and the answer is 1 2 8 7
I got it till here. How did you calculate the coefficient of x 1 0 0 ?
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I've added some details.
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What I mean by symmetry is that the polynomial is Palindromic
P ( x ) = ∑ i = 0 n a i x i is a polynomial of degree n, then P is palindromic if a i = a n − i for i = 0 , 1 , . . . , n
We use the facts that:
-
the product of two Palindromic polynomials is Palindromic polynomial,
-
and power of a Palindromic polynomial is Palindromic polynomial.
1 + x + . . . + x 9 is Palindromic ⟹ ( 1 + x + . . . + x 9 ) 3 is Palindromic
1 + x + . . . + x 2 5 is Palindromic ⟹ ( 1 + x + . . . + x 2 5 ) 3 is Palindromic
And finally ( 1 + x + . . . + x 9 ) 3 ( 1 + x + . . . + x 2 5 ) 3 is Palindromic
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@Abdelhamid Saadi – That's alright. I got that. Thanks! Didn't think about symmetry then.
i still don't get it how can it be symmetry?
1 2 3 4 5 6 7 8 9 10 11 |
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1 |
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Well what's the answer then?
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its hard if you using GF or inclusion exclusion (i have tried). so, i just coding it with c++. the answer is 1287. still waiting for great solution '-'
@Muhammad Ihsan Nice work!
In the future, you can actually display your code directly by copying and pasting it into something like this:
For example:
1 2 3 4 5 6 7 |
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This will allow others to read and copy your code more easily. :)
The highest possible value is $108, for ZZZ 000. Finding the number of license plates that cost $100 dollars is equivalent to finding how many ways 8 indistinguishable balls can be placed in 6 distinguishable boxes, without exclusion. This leads to a value of 1287.