<h1>18. (grid-in)</h1>
<p>in the integer 3,589 the digits are all different and increase from left to right. how many integers between 4000 and 5000 have digits that are all different and that increase from left to right?</p>
<hr>
<p>answer's 10. i had to use a probablility tree to figure it out and took FOREVER. is there anyway to do this problem without using a tree? it seems like a dependent probability problem, but i'm clueless</p>
<p>thanx a bunch.</p>
<p>The first digit has to be a 4. Then, consider any set of three distinct integers between 5 and 9. For every such set, there's exactly one way to order the three numbers in the set so that it increases from left to right. If you know what the last three numbers are, you know there's exactly one way of arranging them. So all you have to count is the number of ways to choose three distinct integers out of 5(5-9). So the answer is 5 choose 3 = 10.</p>
<p>u can narrow the integers down between 4500 to 5000, because anything under that violates the rule. when u start the 4500's, the next number has to 456x , then 456x, and so on. so you just use logic and reasoning to figure out which ones dont violate the rule all the way up to 5000. it took me bout 2 minutes and all i did was use common sense, no probability tree or equation.</p>
<p>^^^ Read what I wrote again... if you choose {6, 7, 9} or {9,6, 7} out of the integers 5-9, you're going to come up with the 4-digit number 4679 either way. Again, given any set of three integers, there's only one way to arrange them in increasing order. If you counted {6, 7, 9} and {9, 6, 7} as two separate solutions, you would be counting the number 4679 twice.</p>
<p>thanx a lot! ur a genius. hope i don't see this on the real thing</p>
<p>Which book had this example?</p>
<p>i just practiced with this same exact test, and i got 14 because i stupidly counted numbers in the 5000s. <em>smacks self</em> </p>
<p>thanks for ruining my self-esteem guys, lol.</p>
<p>this is a QAS. and tnb, i got 9 cuz i forgot a stupid number in my stupid tree.</p>
<p>Can you tell me which QAS? Is it on the googlegroups CC SAT Prep site?</p>
<p>yup. may 20 2006. i think... hold on.. i'll check..</p>
<p>yup.</p>
<p>I tried the May 2007, and like 10 of the questions had important facts scratched out while it was being copied, so I gave up on that one. How's the Math on this one though? I'm going to take it tomorrow if I have time. Are you taking CR on these tests? How do you take them? Do you print out the passages and questions? It takes forever to print since there are so many black marks on the pages though, so I usually don't do CR.</p>
<p>About the problem, I have
4567, 4568, 4569, 4678, 4679, 4789. What are the other 4 again(since you can't have 4890)?</p>
<p>4578, 4579, 4589, 4689, i think those are the four you're missing</p>
<p>Well I guess I'm getting this right tomorrow when I take this test tomorrow lol. The Grid ins are always the most dangerous if you're trying to aim for a very high score, does anyone else feel this way?</p>
<p>Nah, I think grid ins are the easiest math problems. The last Multiple Choice in every section is the hardest because the test-makers know all the little careless (but common) mistakes and use those to make the answer choices agree with your answer no matter what.</p>
<p>I completely agree with amb3r's method.When faced with counting problems it is best to sit back and think an easier way rather making a "tree".Also one should note the distinct difference between combinations and permutations.In the above problem amber used combination not permutation.I hope this helps.</p>
<p>Redwood,you are wrong.The number shouldn't necessarily start with 45.</p>
<p>okay so first digit is 4.</p>
<p>and there is a 1-1 correspondance between</p>
<p>ANY RANDOM UNORDERED 3 DIGITS THAT ARE ALL GREATER THAN 4</p>
<p>and</p>
<p>numbers w/ first digit of 4 AND digits increasing from left to right</p>
<p>because any random 3 digits can be ORDERED to be increasing.</p>
<p>and the only possible digits are: 5, 6, 7, 8, 9 cuz they're all greater than 4</p>
<p>you want to chose 3 digits from 5, so</p>
<p>so you have (5!)/(3!)/(5-3)!</p>
<p>which = 10</p>
<p>honestly this doesn't seem to be SAT Math difficulty...</p>
<p>Narcissa..did you find it easy cuz many people would probably miss it.It is not an easy question.</p>
<p>
[quote]
honestly this doesn't seem to be SAT Math difficulty...
[/quote]
eh that sounds like i thought it was easy.</p>
<p>no i meant this is harder than normal SAT math question</p>
<p>it might take a bit of time to get the answer but i think the bigges tproblem w/ ppl taking SATs is that most people see this and go "wow i've never done something like this before, ithink i don't know how to do it." and people who think like that never usually get the answer.</p>
<p>but i dind't know there were too much combinatorics questions on the SAT</p>