0
$\begingroup$

Find all four digit numbers such that sum of digits is $10$.

If $x_1x_2x_3x_4$ is our number then $x_1+x_2+x_3+x_4=10$. Number of solutions using stars and bars is $C_{13}^{10}$ which is equal to $286$. But we also need to subtract cases when we have $10$ as a digit which we have $4$ cases so answer should be $282.$ But answer is $219$ can you help to understand where is problem?

$\endgroup$
1
  • 2
    $\begingroup$ Did you include $0235$? $\endgroup$ Commented Jul 3, 2022 at 14:42

1 Answer 1

3
$\begingroup$

By four digit numbers, we mean that the first digit can't be $0$ (otherwise it'll be lesser digit number). So $x_1 \geq 1$

So, the answer is the number of non-negative solutions to the equation $$x_1 + x_2 + x_3 + x_4 = 9$$ Why do we subtract $1$? Consider already adding the minimum $1$, so equation becomes $$(x_1 + 1) + x_2 + x_3 + x_4 = 10$$ The number of solutions is $$\binom{9 + 4 - 1}{4 - 1} = 220$$But, we have to subtract $1$ case where the entire $9$ is given to the first digit (so it becomes $10$ which isn't a digit). So, answer = $220 - 1 = \boxed{219}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .