<p>In the integer 3,589 the digits are all different and increase from left to right. How many integers between 4,000 and
5,000 have digits that are all different and that increase from left to right?</p>
<p>Answer is 10.</p>
<p>4567
4568
4569
4578
4579
4589
4678
4679
4689
4789</p>
<p>Yep. 10.</p>
<p>yeah is there another way for solving this ? </p>
<p>I THINK that listing them is the only way to solve this.
Ofcourse that is just what I THINK. lol…</p>
<p>Actually, there is another way, very clever. Check out the first hit that comes up, second post:</p>
<p><a href=“3,589 the digits are all different and increase from left to right site:talk.collegeconfidential.com - Google Search”>3,589 the digits are all different and increase from left to right site:talk.collegeconfidential.com - Google Search;
<p>EDIT: Now, for some reason, that link isn’t working for me. In any case, the idea was this: You know the number has to begin with 4. For the remaining digits, you have to choose 3 from the digits 5,6,7,8,9. 5C3 = 10. You use 5C3 and not 5P3 because no matter what 3 you pick, you have to arrange them in ascending order. </p>
<p>I admit that when I did this problem, I just used brute force. I’ve always admired the more elegant solution.</p>
<p>Because the range is only 4000 to 5000, there really aren’t that many numbers, so listing would be most efficient. However, if the question were to ask between 1000 and 10000, then it would be more efficient to find a pattern and work it out. That kind of question would be more competition-math oriented.</p>
<p>Yes. But then when you go a little further, brute force becomes easier again! As in: Find all of the integers greater than 10,000,000 and less than 100,000,000 with ascending digits. Or between 100,000,000 and 1,000,000,000…but then we have to stop!</p>
<p>@pckeller - could you please explain what 5C3 and 5 P3 mean and how they are used in calculations? Thanks!</p>
<p>Well, you don’t need this for the SAT (where brute force is rewarded!), but let’s see what we can do…</p>
<p>Say you have 5 people: A, B, C, D, and E. And you are selecting from them a president, secretary and treasurer. How many ways can you do that?</p>
<p>The counting principle says that since you have 5 choices for president, then 4 for secretary, then 3 for treasurer, you have 5x4x3=60 ways to do this. This is called 5P3. The P is for “permutation”, a word that means “arrangement”. You have 5 things and you are building arrangements of 3 out of those 5.</p>
<p>When the numbers are bigger, it is helpful to have a formula for permutations.<br>
5P3 = 5x4x3 = 5!/(5-2)!</p>
<p>And in general, nPr = n!/(n-r)!</p>
<p>Now, suppose that instead of selecting officers, you decide to form 3 person committees. Are there still 60 ways? No. The answer 60 counted ABC and ACB as different. But if we were forming committees, they would be the same. In fact, there are 3x2x1=6 ways to arrange those 3 people. So our answer (60) overstated the number of committees by a factor of 6. The answer now is 60/6=10.</p>
<p>We just did (5P3)/(3!) to find the number of committees. That is called 5C3. The c is for “combination”. </p>
<p>In general, to go from nPr to nCr you just divide by r! </p>
<p>That means: nCr = n!/((nr-)!(r!))</p>
<p>Getting back to the original question: we need to know how many ways we can select a trio of digits from the digits 5, 6, 7, 8, 9. Order doesn’t matter because after we select them, we have to list them in ascending order. So it is like choosing a committee of 3 numbers out of these 5 numbers: 5C3 = (5x4x3)/(3x2x1) is what we need.</p>
<p>Like I said, other than the counting principle, you don’t need any of this on the SAT. But it is cool stuff…</p>
<p>Some find it easier to list the possibilities by drawing it as a tree that branches out for each choice. Combinations and Permutations are included in the SAT but not as scary as they sound (<a href=“Bloomberg - Are you a robot?”>Bloomberg - Are you a robot?)</p>
<p>OK, but notice that all of the examples in the article can be solved with nothing more than the counting principle. You can definitely get an 800 w/o knowing nPr, nCr. I am curious whether the new SAT will have more combinatorics or less. </p>
<p>I have a feeling that the new SAT has less combinatorics, which I’m honestly not so sure if I agree with. Judging by the new SAT topics and some sample problems that I saw, the new math section seems very algebra heavy.</p>
<p>Thank you everyone 
and @pckeller thanks so much for you articulate explanation :)</p>
<p><a href=“Box”>https://app.box.com/s/5lq25lhegnhno6e8ujgl</a> Hi please help with number 8 14 . 15 and 16 i feel my answer is right but on the answer key it </p>