<p>I'm not sure if I'm allowed to post certain links but if there is a problem I will remove the link to comply with CC rules.</p>
<p>I'm not too good with combination and permutation problems- does anyone have a simple solution to this problem- The</a> Official SAT Question of the Day</p>
<p>Thanks!</p>
<p>First of all, only 3 get an office. So you don’t care about the cubicles. Make three spaces.</p>
<p>___ ___ ___: Now there are 3 Men (M) and 2 Women (W)</p>
<p>Probability=(WIN)/(TOTAL COMB)</p>
<p>First lets find out total DISTINCT Combinations
5x4x3=60 (5 choices for first space, 4 for 2nd space, 3 for 3rd space). Since its distinct divide by 6 (3*2).</p>
<p>Thus our formula becomes: Probability=(WIN)/10
Now lets just figure out the ways to win, or match 2 men with 1 women.</p>
<p>[1] Assume MMW: x2 MMW/WM or MMW/MW
[2] Assume MWM: x2 MWM/WM or MWM/MW
[3] Assume WMM: x2 " " “” " </p>
<p>6/10=[3/5]</p>
<p>The easiest way is my way.</p>
<p>3/5 = Male
2/5 = Female</p>
<p>2/5 –> Male
1/5 –> Female</p>
<p>2/5 + 1/5 = 3/5</p>
<p>You can’t get easier then this method really. The post above me is the long way to do it.</p>
<p>nothingto- There was nothing to that problem =).</p>
<p>I understand your explanation, but perhaps not the logic behind it- still a bit unclear.</p>
<p>Thanks guys!</p>
<p>Yea I’d be interested in understanding why nothingto’s solution works. Seems a bit too good to be true.</p>
<p>You CAN do it by listing out the possibilites.
Call the men X, Y, Z. Call the women A,B.</p>
<p>List the ways to pick 3…</p>
<p>XYZ XYA XYB XZA XAB XZB YZA YZB YAB ZAB</p>
<p>Count and see that 6 of the 10 contain 2 men, 1 woman.</p>
<p>But I admit that this is a bigger-than-usual pain for these kinds of problems.</p>
<p>You CAN do it using combinatorics/probability. But that requires more knowledge of that subject than is ever actually tested on the SAT. </p>
<p>So my first question is: do the QoD’s usually match the SAT in difficulty? I’m guessing that the answer is often but not always. I don’t follow the questions enough to know their source or the trend in difficulty.</p>
<p>And my second question is for nothingto – Where did you get your second set of probabilities? And why do you add?</p>
<p>^^^
and
^^
also
^</p>
<p>It works because I am awesome. Please save your applause, as I am not much of a show-off. I try to help people as much as I can 
:)</p>
<p>Seriously: I don’t know how this method worked, but it did lol. I must be awesome like that.</p>
<p>I just did this…</p>
<p>Since there is 3 men and 2 women. Since the main area of concern is about the 2 men and 1 women, that was my second probability set. </p>
<p>so </p>
<p>3/5 and 2/5
2/5 and 1/5</p>
<p>I added 2/5 and 1/5 because it asked for the probability of both the 2 men and 1 women. 2/5 and 1/5 is 3/5.</p>
<p>^That won’t always work. Doing it my method is usually best for these types of problems, especially when they get harder. Lets say instead of 5 people, there were 12 - then you have to use combinatorics and can’t count each case. I would recommend learning the mathematical approach to a problem like this. </p>
<p>However, this problem is simple enough where a ‘brunt force’ approach is a decent alternative.</p>
<p>Thanks GreedisGood for your explanation!</p>