I will assume by "number" you mean positive integer.

Break 84 into its prime factors.

84 = 2*2*3*7

I. First we will get all possible 4 digit integers with product of digits 84:

A. Any 4 digit permutation of the digits 2,2,3,7 will have product 84

   There are 2 indistinguishable digits, so the number of distinguishable
   permutations of those is  =  = 12

B. We can multiply factors 2*2 and get the digit 4 and introduce 1 to 
   get 4-digit integers with digits 4,3,7,1

   There are 4! or 24 permutations of these

C. We can multiply factors 2*3 and get the digit 6 and introduce 1 to 
   get 4-digit integers with digits 6,2,7,1

   There are 4! or 24 permutations of these

So there are 12+24+24 = 60 four-digit integers with product of digits 84.

II.  Next we will get all possible 3-digit integers with product of digits 84.

A. We can multiply factors 2*2 and get the digit 4.  So any three-digit 
   integer with digits 4,3,7 will have product of digits 84

   There are 3! or 6 permutations of these

B. We can multiply factors 2*3 and get the digit 6. 
   So any three-digit integers with digits 6,2,7

   There are 3! or 6 permutations of these

So there are 6+6 = 12 three-digit integers with product of digits 84

III.  There are no 2-digit integers with product of digits 84, since the
      largest product of digits is for 99, which has product of digits of
      only 81. 

Answer:  60 four-digit integers and 12 three-digit integers.

         Total 72.

[Note: If you actually mean numbers, and not just integers, you would have
 to count each four-digit number like, say, 2347 five times, as 
 .2345, 2.347, 23.47, 234.7, and 2347.  So there would be 60×5 or 300 four
 digit numbers.  Also each 3-digit number like , say, 437 four times, as 
 .437, 4.37, 43.7, and 437.  So there would be 12×4 or 48 three-digit numbers.
 And in that case the total number would be 348.  However I believe you
 meant just integers.]      

Edwin