Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! = 120.
Now, in half them unit's digit will be bigger than the ten's digit and in half of them it will be smaller.

Example: Let's say we have three digits 1,2,3. Total number of numbers without repeating digits = 3*2*1=6
Numbers with Unit's digit greater than the ten's digit
123, 213, 312
Numbers with Ten's digit greater than the unit's digit
321, 132, 231

So total Number of cases = 120/2 = 60
So, Answer will be B
Hope it helps!
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Does it mean a five digit number? A number can be 2 digit , 3 digit till 5 digit for this combination

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Condi-1:Digit at unit place> digit at tens place.
Condi-2: Without repetition
(1,2,3,4,5)

possible combinations for tens place and unit place, 5C2= 10. Here we will not multiply by 2! because we want ascending order. For example, (2,1) and (1, 2) are two pair but we need only (2,1) which is satisfying condition-1

For remaining places, arrangement of remaining digits is 3*2*1= 6.
So total ways of arrangement= 6*10= 60.
B is answer.