<p>If there are 6 offices that are randomly assigned to 6 people, and 2 of the 6 offices have x's on them, what's the probability that both karri and jone will each be assigned an office with an "x"? I went 4!+4!/6! and got 1/15.... Anyone know a lot about mutual combinations or whatever it's called.</p>
<p>I think your right because if Karri has the first office and Jone the last, then there are four available spaces for the other four, which is 4!. Then muliply it by two, since Jone can be first and Karri the last, so you get 2 x 4!. Then divided this by the total probability of 6! to get 1/15.</p>
<p>correctamundo</p>
<p>Wow! I actually got a problem right that the CB considers to be hard. Write this one down!</p>
<p>What if there are 6 offices, and 3 have X's and we want to know the probability of chris and heather each getting one. How would you solve this? 3!*3!/6! ?</p>
<p>ehh...i usually do problems like this differently.</p>
<p>for your first one, for example, i had C(2,2)/C(6,2). out of 2 possible x's, you need both of them. the total number of possibilities is choosing 2 offices out of 6.</p>
<p>so for the second one, the denominator stays the same, but the numerator becomes C(3,2), because they're are two people but three offices with x's. so the answer would be C(3,2)/C(6,2) which is 3/15 or 1/5.</p>
<p>One way to solve the problem in post#5:</p>
<p>Think of the offices lined up, with the 'x' offices first. Now choose occupants for each office, one at the time. Assignments with Chris & Heather getting one of the 'x' offices are:</p>
<p>CH_ : probability (1/6)(1/5) = 1/30
HC_ : probability (1/6)(1/5) = 1/30
C<em>H : probability (1/6)(4/5)(1/4) = 1/30
H</em>C : probability (1/6)(4/5)(1/4) = 1/30
_CH : probability (4/6)(1/5)(1/4) = 1/30
_HC : probability (4/6)(1/5)(1/4) = 1/30</p>
<p>Sum of probabilities for all such assignments = 6(1/30) = 1/5 .</p>
<p>(Note that this is just one possible explanation; cleaner ones exist)</p>