Choose three numbers from 1, 2, 3, 4, 5, 6, 7, 8, 9 and make their sum odd. How many different choices are there? I heard students say: it seems that there is a formula to calculate this, but I don't know what it is So please help me to be complete. If it's good, add more wealth reward value. (it must be understood by children)!

Choose three numbers from 1, 2, 3, 4, 5, 6, 7, 8, 9 and make their sum odd. How many different choices are there? I heard students say: it seems that there is a formula to calculate this, but I don't know what it is So please help me to be complete. If it's good, add more wealth reward value. (it must be understood by children)!

There are 5 odd numbers {1,3,5,7,9} and 4 even numbers {2,4,6,8}
The sum of three numbers is odd, there are two cases, two even, one odd and three odd
So C (1,5) × C (2,4) + C (3,5) = 35

Use the ten numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 to form five two digit numbers. Each number can only be used once. If the sum of them is odd, then the five two digit numbers can be used

92+73+40+65+81
=351

Take any two numbers from 1,2,3,4,5,6,7,8,9, and the sum of the two numbers is odd
Answer with permutation and combination

The sum of an odd number and an even number is odd. In other cases, both sums are even
The general situation is: C9 (2) = 36 kinds
There are 20 cases of an odd number and an even number: C5 (1) * C4 (1) = 20
The probability that the sum of two numbers is odd is: 20 / 36 = 5 / 9