<p>Well, what if the question asked for 2 democrats and 1 republican instead? Would you still get 3/5?</p>
<p>What if the question asked for 3 republicans?</p>
<p>You can’t just attempt to put numbers together because they “work” once or twice.</p>
<p>It works in this situation because of the specific numbers chosen. It is not asking to pick all of one side. If the question was different I would choose another method.</p>
<p>How about 2 democrats and 1 republican? Does that yield 3/5?</p>
<p>People often get confused from such problems because of the abundance of ways you can explain them. You can count with numbers, letters, whatever. </p>
<p>Learning the combination formula by far is the fastest way to solve these problems.Counting takes way too much time.</p>
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<p>I lol’d so hard after reading that. I’m trying to do his book right now, and it’s taking me around 10 minutes to finish reading and solving one of his 50 tips. lol :)</p>
<p>
I often contemplate the effectiveness of Chung’s. I don’t really see why the book was so highly recommended on CC. It’s super hard, but I guess it makes you look at problems more in depth, such as the one I had posted originally.</p>
<p>I personally regret for not buying Chung’s and Barron’s guide but instead jumping on Gruber’s and Barron’s 2400 (which ,to me, looks like a brusquely cut version of the whole Barron’s guide).
Even though I already attain 760-780 on Math, It took me month and a half to get from 400 to 760.This consists of 6-7 hours daily practice on both math and CR (tried to allot my time equally).</p>
<p>I guess if I bought those two in the first place, It would take me a week or two less to get the desired results (at least for math, CR is much more complex). But it’s too late now … I’ll be attending institutional SAT on 24th Feb…all I can do is practice.</p>
<p>So choose your preparation materials wisely if you’re out of time to prepare. ESPECIALLY if you’re out of time to prepare.</p>
<p>^Thanks for the advice I’ll be taking it in approximately one month.</p>
<p>I’m getting horribly off topic but I just have to know, what is “institutional SAT?”</p>
<p>^He probably means international SAT.</p>
<p>Anyway, this problem from Chung’s is just out of left-field. The trickiest counting problem I’ve seen involved groups of people, and for people, remember that:</p>
<p>Jill and Jack = Jack and Jill. </p>
<p>Therefore, divide by the number of people in a group.</p>
<p>Example: How many ways can you send two of three plumbers to a job (Jill, Jack, and Rob). </p>
<p>3*2 = 6. </p>
<p>6/2 = 3 ways.</p>
<p>^ Well it won’ tbe on the test, but for the record, you don’t divide by the number of people – you divide by the factorial of that number. (It worked for your example b/c 2!=2)</p>
<p>So for Allen, Bob, Cathy, Dave, Edward and Fran, if you want the number of 3-person groups, you need 6x5x4/(3x2x1). But of course, that is exactly what we mean when we say 6C3. </p>
<p>But again: I’ve never seen a problem on the SAT where you needed nCr. When it seems like you need that, the numbers are always small enough to list and count. So from that chapter of your algebra book, the only thing you need is the counting principle.</p>
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<p>Thanks pckeller for the correction. It should read: Therefore, divide by the factorial of the number of the people in a group.</p>
<p>Hope this didn’t cause too much confusion :).</p>
<p>Haha the QOD for today is exactly the same as this and I STILL got it wrong. At least I was close, 1/2 XD</p>
<p>You have to find the total number of possibilities
5C3 = 10 total unique combinations
Then calculate the possibilities of 3:2 committee as the prompt says
3C2 = 3 for the republicans
2C1 = 2 for the democrats
Therefore, such a committee can be composed in (3*2) different ways, or 6.
The final answer is 6/10 or 3/5</p>
<p>^^ It makes sense to read the whole thread before posting. DrSteve (in post #2) already offered this solution.</p>
<p>Compare Dr. Chung’s question
to [The</a> Official SAT Question of the Day](<a href=“SAT Practice and Preparation – SAT Suite | College Board”>SAT Practice and Preparation – SAT Suite | College Board):
Striking similarity!</p>
<p>After forming a subcommittee we have 1 Republican and 1 Democrat left as a counter-subcommittee, therefore the original question is equivalent to
What is the probability of selecting 1 Republican and 1 Democrat for a counter-subcommittee?</p>
<p>This question can be done then the same way xiggi cracked the offices/cubicles question: <a href=“http://talk.collegeconfidential.com/sat-preparation/985813-quick-way-do-problem.html#post11071814[/url]”>http://talk.collegeconfidential.com/sat-preparation/985813-quick-way-do-problem.html#post11071814</a>.</p>
<p>Just making a minimal editing in xiggi’s solution:</p>
<p>Let’s call the Democrats D1 and D2 and the Republicans simply 1, 2, and 3.
It is super easy and fast to see that D1 + 1,2,3 and D2 + 1,2,3 are the only 6 combinations that work, but we need to know the total of all combinations. Thus, the other XX ones.</p>
<p>Possible ONE DEMOCRAT SELECTED
D1 + 1
D1 + 2
D1 + 3
D2 + 1
D2 + 2</p>
<h2>D2 + 3 </h2>
<p>Other possibilities
D1 + D2
1 + 2
1 + 3
2 + 3</p>
<p>The result is 6 over 10.</p>
<p>“No, chung’s. Hence the stupidity.”</p>
<p>^implying that because it’s from Dr. Chung, it can’t be similar to real SAT CB made question</p>
<p><em>finds out that a similar problem was also made by CB Question of the Day with the EXACT same numbers but different wording</em></p>
<p>A blind hog can find an acorn every once in a while. Chung is still mostly irrelevant and not a great choice.</p>
<p>The OP’s question’s not too difficult. There are 3C2 * 2C1 = 6 ways to select a committee with 2 republicans and 1 democrat, and 5C3 = 10 ways to select a 3-person committee. Ans is 6/10 = 3/5.</p>