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Question and my solution is in the paint document.
Case 1  Case 3 represents all possible combinations of string of length 4 with exactly three 9's.
The d represents a digit that is not 9.
d = 0 or 1 or 2 or 3 or ... or 8: This is 9 options.
Notice that none of the cases 1  3 have identical strings because,
Case 1 cannot be equal to any of the strings in case 2 or 3 because Case 1 start with a 9 where case 2 starts with d which cannot be a 9 also case 1 ends with a 9 where case 3 ends with a d which cannot be a 9. Finally the third column of case 2 is a d and the third column of case 3 is a 9.
Therefore each case has 9 distinct strings with exactly three 9's and 9(3) = 27.
The book solution is
9 + 9 + 9 + 9 = 36.
Book solution is in paint doc.
Case 1  Case 3 represents all possible combinations of string of length 4 with exactly three 9's.
The d represents a digit that is not 9.
d = 0 or 1 or 2 or 3 or ... or 8: This is 9 options.
Notice that none of the cases 1  3 have identical strings because,
Case 1 cannot be equal to any of the strings in case 2 or 3 because Case 1 start with a 9 where case 2 starts with d which cannot be a 9 also case 1 ends with a 9 where case 3 ends with a d which cannot be a 9. Finally the third column of case 2 is a d and the third column of case 3 is a 9.
Therefore each case has 9 distinct strings with exactly three 9's and 9(3) = 27.
The book solution is
9 + 9 + 9 + 9 = 36.
Book solution is in paint doc.
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