0
$\begingroup$

How many $5$ digit numbers are possible having sum of their digits = $22$ ?

How can I solve it by using Stars and Bars problem ?

$\endgroup$
  • $\begingroup$ Observe that: $5+5+5+5+5=25$ $\endgroup$ – Bumblebee Dec 10 '16 at 19:27
4
$\begingroup$

This is an inclusion-exclusion question.

Count how many solutions to $a_1+a_2+a_3+a_4+a_5=22$ with $a_1\geq 1$ and $a_2,a_3,a_4,a_5\geq 0$.

Then subtract the number of solutions with $a_1\geq 10,$ then the number of solutions with $a_2\geq 10,$ etc.

Then add back in the cases where two of the $a_i$ are $\geq 10$, since you've subtracted these cases twice.

Each of the values you need for this can be computed with stars and bars.

The ultimate count you get is:

$$\binom{21+4}{4} - \binom{12+4}{4}-\binom{4}{1}\binom{11+4}{4} + \binom{4}{1}\binom{2+4}{4} +\binom{4}{2}\binom{1+4}{4}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.