How many 10 digit multiples of 99 are pandigital? Remember, no leading zeros.
(In reply to
solution by Charlie)
Both the explanation and the result are unquestionably correct.
1. However the terms even digits and odd digits are misleading, clearly the intention was even-positioned and odd-positioned.
2. One can get a rough order of magnitude of the result by evaluating
(10!-9!)/11 =about 296,900. The calculation is fast, but the flop side is lack of error's estimate.
re: solution 2 remarks
