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## Discussion of Problem 1044. Lucky Tickets. Easy!

A combinatorial approach
Posted by abid1729 25 Jun 2019 00:31
firstly, if 1st 2 digits and 2nd 2 digits are same then there will be 10*10=100 combinations of 2 digits.
for every number there will be 2 combination.(example:1212 and 1221)
so,combinations will be 2*100=200. But there are such 10 numbers for which 2 combinations are not available.(ex: 0000 , 1111 , 5555)
COMBINATIONS = 200-10 = 190.
now, we have to check from highest 9+9=18 to lowest 0+0=0 that how much sets have same sum.
16<<<< (9+7),(8+8)
15<<<< (9+6),(8+7)
14<<<< (9+5),(8+6),(7+7)
13<<<< (9+4),(8+5),(7+6)
12<<<< (9+3),(8+4),(7+5),(6+6)
11<<<< (9+2),(8+3),(7+4),(6+5)
10<<<< (9+1),(8+2),(7+3),(6+4),(5+5)
9<<<< (9+0),(8+1),(7+2),(6+3),(5+4)
8<<<< (8+0),(7+1),(6+2),(5+3),(4+4)
7<<<< (7+0),(6+1),(5+2),(4+3)
6<<<< (6+0),(5+1),(4+2),(3+3)
5<<<< (5+0),(4+1),(3+2)
4<<<< (4+0),(3+1),(2+2)
3<<<< (3+0),(2+1)
2<<<< (2+0),(1+1)

combinations =(2*2*2/2)+(2*2*2)+(2*2*2+2*2+2*2) +(3*2*2*2)+..................................+(2*2*2)+(2*2) =480

SO, total combinations = 190+480 =670