1. ## Probability question

I have 4 assignments to pass to three different people to do for me. What is the probability that each person will receive at least one assignment?

Total combination: $\displaystyle {4+3-1\choose 3-1}$ = $\displaystyle {6\choose 2}$ = 15

P(each person receive at least one assignment) = $\displaystyle 3\over 15$ = $\displaystyle 1\over 5$

but it's wrong

It's a MCQ question with options: (A) $\displaystyle 8\over 9$, (B) $\displaystyle 64\over 81$, (C) $\displaystyle 4\over 9$, (D) $\displaystyle 16\over 81$, (E) $\displaystyle 5\over 9$

2. Originally Posted by noob mathematician
I have 4 assignments to pass to three different people to do for me. What is the probability that each person will receive at least one assignment? It's a MCQ question with options: (A) $\displaystyle 8\over 9$, (B) $\displaystyle 64\over 81$, (C) $\displaystyle \color{blue}4\over 9$, (D) $\displaystyle 16\over 81$, (E) $\displaystyle 5\over 9$
Consider that the 4 assignments are all distinct.
There are 36 surjective (onto) functions from a set of 4 to a set of 3.
There are $\displaystyle 3^4$ functions from a set of 4 to a set of 3.

Ya if the assignments are distinct, there will be a total of $\displaystyle 3^4$ possibilities.

But how do you find the 36?

My interpretation will be to permute 4 assignments among the 3 people (since each of them will receive one assignment), which is 4. 3. 2. 3 = 72, since the last assignment could be allocated to any one among the three. Therefore answer is A?

4. Originally Posted by noob mathematician
But how do you find the 36?
The number of mappings from {1,2,3,4} to {a,b,c} which are onto is 36.
Onto means that each of a, b, & c gets at least one.
Look up the ways to count surjections.

5. oic. Thanks a lot.

The formula S(n,k)= $\displaystyle \sum_{i=1}^{n} {k\choose i} (k-i)^n (-1)^i$

Ok!

6. Originally Posted by noob mathematician
The formula S(n,k)= $\displaystyle \sum_{i=1}^{n} {k\choose i}(k-i)^n (-1)^i$
That is close. But not exact.
The number of surjections from a set of N elements to a set of K elements is:
$\displaystyle S(N,K) = \sum\limits_{j = 0}^K {\left( { - 1} \right)^j { K \choose j} \left( {K - j} \right)^N }$.

7. Oh ya.. haha some typo error there. Thanks for pointing out.