<p>Yesterday's question seemed to be extremely confusing... and frankly i didn't quite understand their solution. can someone explain this in easier terms.</p>
<p>Of 5 employees, three 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?
A. click to choose answer A 1 over 3
B. click to choose answer B 2 over 5
C. click to choose answer C 1 over 2
D. click to choose answer D 3 over 5
E. click to choose answer E 2 over 3</p>
<p>Why are so many people having trouble with this question? Only took me about 20 seconds to solve. </p>
<p>Anyways, here is how I did it.</p>
<p>6 Distinct ways for this to occur. MMW/MW, MMW/WM, etc. So 6 ways. 10 Distinct combinations total (5<em>4</em>3)/(3*2)=10. 6/10 or 3/5</p>
<p>[D]</p>
<p>This seems to be a troublesome problem if CB officials are lurking this site- don’t add such questions to the SAT!</p>
<p>care to elaborate more Greedisgood?</p>
<p>For 2 men and 1 woman to be in a cubicle the only possible combinations are:</p>
<p>MMW x2 (-MW or -WM left over)
MWM x2
WMM x2</p>
<p>So 6 combinations. </p>
<p>Why just 10 DISTINCT possibilities? </p>
<p>Well, normally there would be 60 (5<em>4</em>3), but some of these are redundant (Remember, out of the 6 spots there are only 2 distinguishable catagories, men and women) So this number (60) needs to be divided. The formula is just the number of each catagory multiplyed by each other. In this case, there are 3 men and 2 women so just divide by 3*2=6. 60/6=10</p>
<p>Probability = 6/10 = 3/5</p>