How many number can we find , that their sum of digits is equal to 6 ?
We have 3 field _ _ _;
We know that the first number cannot be 0 and has to be lower than 7 otherwise its sum wouldnt equal 6.
We are left with two field _ _ . Those numbers also has to be lower than 7 but can be 0. What do we use? Variaions combinations or permutation?
I calculated it using combinatnions using
$\frac{k+n-1}{n} = \frac{ 6 + 2 - 1}{ 2 } = \frac{7!}{(7-2)!*2!} = 21$
Is is this correct? I am still not sure when i should use combination with or w/e repeatition or something else. Could anybody explain the content of when to use what in more depth than basic?
All help is appreciated. Thanks