Permutation and combination question. Whats wrong in my attempt.

My Question: 6 different books are to be distributed among three peoples {R/S/G} . Then number of ways of distribution of books such that each person gets atleast one book.

My attempt:

$$\binom{6}{3} 3! 3^3 = 3240$$

because we can choose first three books with $\binom{6}{3}$ and give them to three people with $3!$. Then we can distribute remaining three books with $27$ different ways.

• Please format your math expressions. – John Dec 20 '17 at 17:40
• Apply inclusion-exclusion to the cases where one or more of the people didn't get any books to arrive at an answer of $3^6-2^6-2^6-2^6+1^6+1^6+1^6-0^6=540$, same as below. – JMoravitz Dec 20 '17 at 18:52

Let $A$ and $B$ be sets such that $|A| = 6$ and $|B| = 3$. How many surjections $A \rightarrow B$ are there?
When number of surjections is asked, you should use "Stirling numbers of the second kind" in order to solve the problem. Then answer of this question is $$3!S(6,3) = 6*90 = 540$$
Here, the number $S(6,3)$ is the number of ways of distributing $6$ different books to $3$ identical people (I don't know how can something like this be possible but you can consider it this way). Since the people are different, you can change the order with $3!$ so the result follows.
Now the problem with your solution is assume you have books $1,2,3,4,5,6$. Then in the first choose, suppose you choose $1,2,3$ and give it them to $R,S,G$ respectively. Then you are distributing the remaining books, suppose $4,5,6$ to $R,S,G$, respectively, similar to the first distribution. But this is same as choosing the first three books as $4,5,6$ and distributing the remaining three as $1,2,3$, giving each people two books. So you are over-counting the possibilities. I also suggest you to learn the Stirling numbers of the second kind and its relation with number of surjections (It is about Inclusion-Exclusion Principle).