Dixon wrote:

I downloaded a file with some sample combinatorics and prob problems, and trying to solve this one:

There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

The answers sheet says it's 8/9. I get 1/5. Here's how I calculate this:

number of possible arrangements:

sec1 sec2 sec3

4 0 0 => 3 arrangements (3!/2!)

3 1 0 => 6 arrangements (3!)

2 2 0 => 3 arrangements

2 1 1 => 3 arrangements, and the only condition satisfying at least 1 report/secretary

arrangements total: 3+6+3+3=15

arrangements where each gets at least one report: 3

Probability of last arrangement: 3/15=1/5

So where am I going wrong? Or is the known answer for this problem wrong?

The reason you are not getting the correct answer is that you are assuming the letters are identical.

You say:

sec1 sec2 sec3

3 1 0 => 6 arrangements (3!) (

Here, you get 6 arrangements becoz sec1 could have 3 or 1 or 0 letters and so on but which 3 letters does sec1 have? The 4 depts will have 4 different letters)

Instead, try to solve it like this:

Total no of ways of distributing 4 letters among 3 secretaries = 3*3*3*3 = 81 (such that some secretaries may get no letters)

To ensure that each secretary should get at least 1 letter, the distribution will be 2, 1, 1.

Select 2 letters out of the 4 and select one secretary out of the three to give the 2 letters => 4C2 * 3C1

Now, you have 2 letters to give to 2 secretaries => 2!

Number of favorable ways = (4*3/2) * 3 * 2

Required probability = 36/81

Try out my blog. I have discussed grouping identical/distinct items in various ways on it.

http://www.veritasprep.com/blog/categor ... om/page/2/Scroll down to the last post. It is the first post on Combinatorics. The 10-12 posts above it are all on probability/combinatorics

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