You mean that the numbers must contain at least the digits $0$, $1$ and $A$ but may contain other hexadecimal digits?
If it's the case, indeed I didn't understand clearly the statement of the problem.
Thanks for your reply. I will check again the statement.
Problem 162
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PierrotLeFou
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PierrotLeFou
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Re: Problem 162
This will be a little more difficult than the previous solution.
Maybe I go in the wrong direction but a part of the solution is to evaluate the possibility to insert the requested digits among the others.
Suppose I have $4$ positions left and $2$ digits to insert among the others.
Here what I found:
$.X.X.$ gives $2X0X0$, $1X1X0$, $1X0X1$, $0X2X0$, $0X1X1$, $0X0X2$.
The$.$ are the possible insertion points and $X$ are the other digits, and the numbers are the amount of requested digits inserted at each position.
I will have to count the number of permutations between inserted digits and possible permutations between those digits and the digits in the $X$ position.
Maybe I go in the wrong direction but a part of the solution is to evaluate the possibility to insert the requested digits among the others.
Suppose I have $4$ positions left and $2$ digits to insert among the others.
Here what I found:
$.X.X.$ gives $2X0X0$, $1X1X0$, $1X0X1$, $0X2X0$, $0X1X1$, $0X0X2$.
The$.$ are the possible insertion points and $X$ are the other digits, and the numbers are the amount of requested digits inserted at each position.
I will have to count the number of permutations between inserted digits and possible permutations between those digits and the digits in the $X$ position.
I'm always right ... until I'm wrong