Powerless power
Let be three distinct one-digit integers.
How many different combinations of triples such that is a one-digit integer?
The answer is 330.
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1 solution
when a is 0 or 1,b and c can be anything as 0 and 1 raised to the power anything is 0 or 1(except 0^0) but as the numbers are distinct this wont create a problem. therefore for base 0 or 1 no. of ways=2x9x8=144 for all other digits as a, the power can be 0 or 1 for a single digit integer so no. of ways=8x23=184(23 because for b as 0 c has 8 possibilities, for b as 1 c has 8 possibilities and for any b and c as 0 there are more 7 possibilities as 1^0 is repeated) for 2^2 no ways are possible for 2^3 and 3^2 one way each is possible so total no.of ways=144+184+2=330