# Solving permutation problem

I need to know the proper way to solve this kind of problem:Five employees of a firm are ranked from 1 to 5 based on their ability to program a computer.Three of these employees are selected to fill equivalent programming jobs.If all possible choices of three (out of the five) are equally likely,then how can we find the following probabilities: a) The employee ranked number 1 is selected. b) The highest-ranked employee among those selected has rank 2 or lower. c) The employees ranked 4 and 5 are selected.

The correct answers are as follows: 6/10 or (36/60) ,4/10, 3/5

I am aware of that the formula for probability is s/n;where n is the total number of outcomes. I know how to get n by using $5_{P_3}$ (5 ways taking 3 at a time). Now, how do I get s (success outcome)? That confuses me. (I know that s(success outcome) is the number that will lock / target your condition) Thank you.

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Your line of thought is a little wrong.

You are choosing $3$ people out of $5$. This means, the order of choice doen't matter. More specifically, choosing $1$ then $2$ then $3$ is the same as $2$ then $1$ then $3$ and so on....

Therefore, the total number of ways of selecting $3$ people out of $5$, is the binomial coefficient $\binom{5}{3}=^5C_3$

The answer to (a) is, since the first guy is always chosen, you have a handle only on the choice of the other two. So, the number of possible ways of selecting $2$ people out $4$ is $^4C_2$

So, the probability is $\dfrac{\binom{4}{2}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}$ which completes (a).

Along the same lines, (c) is also solved this way: You have handle only on one guy which means, you have to choose $1$ out of $3$ guys, So, the probability is $\dfrac{\binom 3 1}{\binom 5 3}=\frac{3}{10}$

For (b), the condition imposed forbids you from choosing the candidate ranked (1). So, You need to choose $3$ out of $4$ people. So, the answer is $\dfrac{\binom 4 3}{\binom 5 3}=\frac{4}{10}=\frac{2}{5}$

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but, shouldn't we use permutations here? coz ranking is invovled (like in the first sentence) ? tnx –  IvanMatala Jan 22 '12 at 10:42
No, did you read my explanation? The point is , three people are chosen; it is like this: choosing $123$ is no different from choosing $231$ and so on... –  user21436 Jan 22 '12 at 10:51
ok. its 5C3, but how do we get "s" (success outcomes) –  IvanMatala Jan 22 '12 at 10:53
how do we get 6 mathematically? i cant figure out / find how to lock down the target –  IvanMatala Jan 22 '12 at 10:55
thanks! i guess i was thinking too much. you made my day –  IvanMatala Jan 22 '12 at 10:59