Question 134503
All we have to know is how many ways can 2 (the middle 2) out of 9 numbers be arranged. The order is important so we do a permutation.
nPr=n!/(n-r)!
=9!/(9-2)!
=9*8*7!/7!
=72
72+9 (for the cases where the digit may appear twice)
=81 
So, there are 81 four digit numbers starting with 1 and ending with 5.
1115, 1125, 1215, 1225, 1135, 1315, 1325, 1235, 1335, only 72 more to go!
.
Ed