probability

The key is to check the last digits.

Other wise also it's simple....See a² leaves a remainder of either 1 or 4 when divided by 5.

So (a⁴-b⁴)=(a²+b²)(a²-b²). Now for the 2nd term to be divisible by 5, a can be of the form 5k+2 or 5k+3. Similarly b must be of the form 5k+1 only.

Similar argument for the 1st term.

Now comes the fun, what when both are divisble by 5.....that happens only when both a and b are divisible by 5 - that completes it - I hope. [124]

Soumik...I tried that reasoning of yours..I was getting a weird answer. Last digits of 4th powers are mostly 6 and 1, so difference of 6-1 gives last digit 5. To give a positive result of a⁴ - b⁴, I discarded some possibilities and got 3 permutations out of 9! possible(because there are digits 0-9 which can be raised to 4th power to check their last digits) ones. My answer is wrong however...maybe I reasoned wrongly.

I got 5/24 for 5 multiples of 5 present in the set of 24 numbers. So 2 x 5/24 (possible for a and b). Dunno, I lost my way after that :D.

did u try out my 2nd argument ?

anyways, I did not calculate it, so maybe there's an error! [12]