combinatorics. how many options to make a ticket?

Question

The company is selling tickets for a show. As they want entice more people to buy a ticket, a company decided to make 10000 ticket which numbers are from 0000 to 9999. If you buy a ticket, where the sum of the first two digits equals to the sum of third and fourth digits, you win a pencil. How many pencils the company should have?

Answer 1

The sum of two digits is a number between 0 and 18. You can easily show that there is:

• 1 combination of digits to create a sum of 18 (99)
• 2 combinations of digits to create a sum of 17 (98, 87)
• 3 combinations of digits to create a sum of 16 (97, 88, 79)
• ...
• 9 combinations of digits to create a sum of 10 (91, 82, ..., 28, 19)
• 10 combinations of digits to create a sum of 9 (90, 81, ..., 18, 09)
• 9 combinations of digits to create a sum of 8 (80, 71, ..., 17, 08)
• ...
• 2 combinations of digits to create a sum of 1 (10, 01)
• 1 combination of digits to create a sum of 0 (00)

Each combination of the first two digits can be matched with any combination of the other two digits with the same sum. For example, if the sum of two digits is 8 you have 9 choices for the first two digits and 9 choices for the last 2 digits which is $9\cdot9=81$ four-digit numbers satisfying the critera.

The total number of tickets is therefore:

$$1\cdot1 + 2\cdot2 + ... + 9\cdot9 + 10\cdot10 + 9\cdot9 + ... + 2\cdot2 + 1\cdot1 = 100 + 2\sum_{n=1}^{9}{n^2}=580$$