1
$\begingroup$

Let $S$ be a set with $|S| = 30$ and let $\pi = \{A_i\}_{i=1}^3$ be a partition of $S$ such that each set $A_i$ of $\pi$ has ten elements. How many such partitions $\pi$ are there?

This questions seems like a deceptively easy question, i.e., just a combination problem. Pick ${30\choose 10}$ for $A_1$, ${30-10\choose 10}$ for $A_2$, and ${30-20\choose 10}$ for $A_3$ Then we have ${30\choose 10}{20\choose 10}{10\choose 10}$. But we don't care in what order the 3 partitions are so finally answer is $$\frac{{30\choose 10}{20\choose 10}{10\choose 10}} {3!}.$$ Is this a sufficient answer?

$\endgroup$
4
  • 3
    $\begingroup$ Yeah, it is ok. $\endgroup$
    – Aqua
    Oct 13 at 20:39
  • 1
    $\begingroup$ From the tag description for solution-verification: "This should not be the only tag for a question." $\endgroup$
    – Shaun
    Oct 13 at 20:41
  • $\begingroup$ @Shaun, I'll fix it. $\endgroup$
    – Seong
    Oct 13 at 20:41
  • $\begingroup$ Thank you, @Owen. $\endgroup$
    – Shaun
    Oct 13 at 20:41
1
$\begingroup$

Your answer is correct. You can also write this in terms of a multinomial coefficient: $$\frac{\binom{30}{10,10,10}}{3!} = \frac{30!/10!^3}{3!}$$

$\endgroup$

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.