<p>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?</p>
<p>This may be wrong…
There are three possible ways of putting two men in an office of three people. There are 5<em>4</em>3 ways of putting people in the office total. so 3/(3<em>4</em>5)=1/20</p>
<p>^Thats incorrect.</p>
<p>There are (5<em>4</em>3) ways of putting the people in the office, but this isn’t distinct. There are (5<em>4</em>3)/(3*2)=10 distinct ways.</p>
<p>There are 6 distinct ways for this situation to occur. Not 3. (MMW/WM, MMW/MW are two distinct)</p>
<p>Thus P=6/10 or [3/5]</p>
<p>^@GreedisGood, Could you explain your answer step by step? I would really appreciate it! Thanks</p>
<p>EDIT: Ive always been horrible with permutations and combinations -_-</p>
<p>For this type of problems, you can use the formula approach, or decide to quickly list the possibilities and use the reasonng approach. After all, this is the SAT!</p>
<p>First realize that it is a bit simpler to look at the cubicle possibilities. If one woman is in an office, only one can be in a cublicle! Let’s call the women W1 and W2 and the men simply 1, 2, and 3. </p>
<p>It is super easy and fast to see that W1 + 1,2,3 and W2 + 1,2,3 are the only 6 combinations that work, but we need to know the total of all combinations. Thus, the other XXX ones.</p>
<p>Possible ONE WOMAN IN A CUBLICLE
W1 + 1
W1 + 2
W1 + 3
W2 + 1
W2 + 2</p>
<h2>W2 + 3 </h2>
<p>Other possibilities
W1 + W2
1 + 2
1 + 3
2 + 3</p>
<p>The result is 6 over 10.</p>
<p>wow thanks xiggi, I didnt even think of the reasoning approach as I’m more of a formula-oriented person. It’s so effective!</p>
<p>Thanks xiggi!!</p>
<p>nice xiggi</p>
<p>For those who want to know the formula method (which would be useful if big numbers like 100 are used), this is how it works:</p>
<p>From the first sentence, we know that [o]f 5 employees, 3 are to be assigned an office… This means that we are to choose 3 out of 5, i.e 5C3 = 10. (this represents the total possibilities for the 3 to be chosen)</p>
<p>3 of the employees are men and 2 are women combined with offices will be assigned to 2 of the men and 1 of the women means you are to choose 2 out of the 3 men and 1 out of the 2 women:</p>
<p>3C2 x 2C1= 3x2 =6. (you multiply because of the word and; this represents how the 2 men and 1 woman are to be chosen).</p>
<p>Therefore, the probability of the 2 men and 1 woman is 6/10 or 3/5. </p>
<p>For a brief recap on combinations, refer to [this</a> link.](<a href=“File Not Found [404 Error]”>File Not Found [404 Error])</p>
<p>Oh damn, I did permuations instead of combinations 
My bad, sorry</p>
<p>quick way=quick mistake</p>
<p>^</p>
<p>That is not necessarily true.</p>