0
$\begingroup$

Assume $n \geqslant k$

What's the possibility? Example value $n = 12$, $m = 15$, $k = 11$

The context is that there are 12 Chinese zodiacs. Someone mentioned to me that out of 15 of his family members, they found 11 different zodiacs. Only one left. What's the odd!

I understand that all zodiacs are not equally likely, but for simplicity let's assume so.

I understand that if $n = m = k$, then $P = n!/n^n$. The first flower can pick any of the n colors, the second flower can pick from the rest n-1 colors, and so on. So it's $$\frac{n}{n}\times\frac{n−1}{n}\times\frac{n−2}{n}\times...\frac{1}{n}=\frac{n!}{n^n}$$

How about the general case? Or at least the example above?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

EDIT : I was completely wrong. Now I edited.

Note that each of $m$ flowers has $n$ possibilities about its color.

Now, first we decide $k$ colors to choose, and second choose at least one flower in each color.

The last step is represented as $$\frac{\{(k-1)+(m-k)\}!}{(k-1)!(m-k)!}=\binom{m-1}{k-1}.$$

However, the answer is not $$\frac{\binom{n}{k}\cdot \binom{m-1}{k-1}}{n^m}.$$ This is because I also have to think about the arrangement of each set...

To be honest, I don't know how to get it.

$\endgroup$
11
  • $\begingroup$ I think the denominator should be 12^15 since there are 12 choices. $\endgroup$ Jan 3, 2014 at 17:13
  • $\begingroup$ @SteveODonnell: Oh, you are right. a big mistake...thanks for pointing it out. $\endgroup$
    – mathlove
    Jan 3, 2014 at 17:20
  • $\begingroup$ This does not match OP when $m=n=k$ $\endgroup$
    – Empy2
    Jan 3, 2014 at 17:30
  • $\begingroup$ @Michael: Thanks for pointing it out. I hope this time is fine. $\endgroup$
    – mathlove
    Jan 3, 2014 at 17:59
  • $\begingroup$ In my post, I think if $n=m=k$, then $P=n!/n^n$ because the first flower can pick any of the n colors, the second flower can pick from the rest n-1 colors, and so on. So it's $$\frac{n}{n} \times \frac{n-1}{n} \times \frac{n-2}{n} ... \frac{1}{n} = \frac{n!}{n^n}$$ $\endgroup$
    – Boyang
    Jan 4, 2014 at 1:45

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.