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I'm absolutely terrible at calculating these things and I would like to, especially with SATs coming up, improve my capabilities.

What always gets me is that there are so many types of ways to combine items.

One way might be to figure out rearrangements of a set:

abcd => abcd abdc acbd acdb adbc adcb bacd badc...

...or combinations of the elements of the set used more than once

abcd => aaaa aaab aaac aaad aaba aabb ...

...or combinations limited to two elements

abcd => aa ab ac ad ba bb bc bd ca cb cc...

...or in threes

abcd => aaa aab aac aad aba abb abc abd...

...or ways to combine the elements in pairs

abcd => ab ac ad ba bc bd ca cb cd da...

...or pairs without repeating

abcd => ab ac ad bc bd...

Point taken, I'm sure. I know they're all incredibly simple: multiply two numbers or find the summation and then divide by the number or times it's going to repeat... Again though, I have no idea which calculations to apply to which variations. I know it should be logical, but I can never quite figure it out.

And then there was an SAT practice problem online... http://sat.collegeboard.org/practice/sat-question-of-the-day?questionId=20120221&oq=1

Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

I don't even know where to begin to solve this problem..

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3 Answers 3

up vote 4 down vote accepted

There are $\binom{5}{3}$ ways to choose the $3$ people for the offices, all equally likely. There are $\binom{3}{2}\binom{2}{1}$ ways to choose $2$ men from the $3$ available, and $1$ woman from the $2$ available. So the required probability is $$\frac{\binom{3}{2}\binom{2}{1}}{\binom{5}{3}}.$$

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right answer is 3/5 yes @Andre Nicolas ? –  dato datuashvili Mar 24 '12 at 6:42
    
Yes, that is the right number. –  André Nicolas Mar 24 '12 at 7:00

So, for all of the various combinations and rearrangements, there are only three things to keep in mind: how many items do you have to choose from, how many choices do you have to make, and does the order you choose them in matter (i.e., is ABCD the same or different from DCBA).

To start, always figure out all the possible permutations (if order doesn't matter we'll need to divide out all of the "same" arrangements at the end). This is done by just multiplying the number of objects you have to choose from by the number of choices you have to make. So, picking $4$ objects from A,B,C,D without repetition, you have $4$ choices, then $3$, etc, so you have $$4*3*2*1 = 4! = 24$$ arrangements. If you can repeat, then it is just $$4*4*4*4 = 4^4 = 256$$ (since you have $4$ choices at each step since they don't get used up).

Then, if the different arrangements of the same letters are considered the same thing (i.e., ABCD = DCBA), we need to divide by the number of times each choice is getting counted. So, if we choose $3$ objects, there are $3*2*1 = 3! = 6$ different ways to arrange them. So we would start out with $4 * 3 * 2 = 24$ permutations, but we need to divide by $6$ to get rid of all the repetitions, so we find there are only $4$ distinct ways to do this. Does that all make sense? If the order doesn't matter part is confusing, try writing down all the combinations for some manageable set and look at which ones are distinct.

Now the SAT question:

We are choosing $3$ office workers out of the $5$. How many ways can we do this? From above, we know we can do this $5*4*3 = 60$ different ways ($5$ objects to choose from, $3$ choices to make). Since we are only concerned with there being $2$ males and $1$ female, the order we choose them in doesn't matter. (MMF is just as good as MFM.) So for each choice of $3$ workers, we will have counted it $6$ times (this just the number of ways to arrange those $3$ objects, which is $3*2*1$). So that means, we have $10$ distinct ways to choose $3$ workers to get an office. Now, out of this ten, how many contain $2$ men and $1$ woman? We apply our principles from above again, this time we have $3$ men from which we need to choose $2$. So $3*2 = 6$, but the order still doesn't matter (Male 1 and then Male 2 is the same as Male 2 and then Male 1), so have to divide by $2 * 1 = 2$. So we have $3$ ways to choose $2$ males out of the $3$. Obviously, there are $2$ ways to choose $1$ woman from the $2$ women. Hence, we have $6$ ways to choose $2$ Men and $1$ woman, so $6/10 = 3/5$ probability.

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totaly possibility for choosing 3 worker out of 5 is 5*4*3=60, from this 60 posibilities,3 distinct group can be choosen in 6 ways,three ways for one worker,two for second and 1 for last one,so totally 60/6=10 different ways for 3 workers chosen from 5 workers. How many of these 10 possible groups of 3 workers consist of 2 men and 1 woman? From the 3 male workers, 2 can be chosen in 3 different ways. There are 2 possibilities for the female worker. Therefore, 3*2=6 of the groups of 3 workers consist of 2 men and 1 woman. so it would be 6/10=3/5

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