How many different ways to distribute $30$ different books to $3$ people, such that the number of books given to each of them create an arithmetic series?
So, I thought one of them should have exactly $10$ books, and the others can partition rest of the books however they like.
Example: $1 – 10 – 19$
Choose $10$ books, $\binom{30}{10}$, any of them can have exactly $10$ books, so multiply it by $3$, also others can partition the books however they like, which is $2^{20}$.
In total $\binom{30}{10} \times 3 \times 2^{20}$.
Then, I guess I should be done here. Except, when all of them have $10$ books, so $10-10-10 $ right? Therefore I should discard $2 \times \binom{30}{10} \times \binom{20}{10}$.
In conclusion, $\ 3 \times 2^{20} \times \binom{30}{10} – (2 \times \binom{30}{10} \times \binom{20}{10})$
Is my answer true? If you have an another approach, can you share? Thank you in advance.
Best Answer
It is better to treat the $10-10-10$ case completely separately from the rest. The number of ways out of $2^{20}$ for the other two people to get $10$ books is $\binom{20}{10}$. The number of ways to distribute the books equally is the multinomial coefficient $\binom{30}{10,10,10}=\binom{30}{20}\binom{20}{10}$. Thus the answer is $3\binom{30}{10}\left(2^{20}-\binom{20}{10}\right)+\binom{30}{10,10,10}$, which matches your answer.