<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>Does anybody have an easy solution? My daughter would like to know the shortcut way but doesn't really understand CollegeBoard's explanation. Thanks!</p>
<p>Answer is 10.</p>
<p>4567</p>
<p>4568</p>
<p>4569</p>
<p>4578</p>
<p>4579</p>
<p>4589</p>
<p>4678</p>
<p>4679</p>
<p>4689</p>
<p>4789</p>
<p>^ sums it pretty well. I guess the key thing here is to not panic when you see that you have 1000 numbers to choose from. I don’t think there is a clear cut way to approach this problem other than realizing that the first digit is 4, so the second must be greater than 4, that is, it has to be 5,6,7,8, or 9. And based on that, you can expand out a combination of numbers, given by Seachai above.</p>
<p>There is a clear-cut way, but I would not say that it is preferable. I believe that, for all SAT questions dealing with combinations, permutations, probability, etc., this method is unnecessary–i.e., you can bang them all out just like the first poster did. I am going to explain a more mathematical way of doing this. Please do not feel obliged to read or to learn this. I mean it. It isn’t necessary. But I’m going to explain it anyway in case anyone is curious:</p>
<p>Since there are 5 choices (the first number must be 4, so we have 5, 6, 7, 8, and 9) and 3 available slots (the hundreds, tens, and ones place), we do 5C3 (“5 choose 3”). The value of 5C3, which is 10, refers to the number of combinations possible for n=5 items to be taken r=3 items at a time. (nCr is a function on a typical scientific/graphing calculator, such as the TI-83; you input the first number, 5, which represents n, activate MATH > PRB > nCr, and then plug in the second number, 3, which represents r. It should look like this: “5 nCr 3.” Press ENTER and you’ll see that the answer is 10. You can also do it algebraically by using the equation n!/[(n-r)! * r!]. 5C3 = 5!/[(5-3)! * 3!] = (5<em>4</em>3<em>2</em>1)/[(2<em>1)</em>(3<em>2</em>1)] = 120/12 = 10.)</p>
<p>The reason we do 5C3 (combination) and not 5P3 (permutation) is because order matters. If we have abcde (5 letters), and we are given a set of 3 letters, {a, b, c}, we can only arrange them one way: abc; bac, acb, cba, and cab do not count. Similarly, for the original problem we were talking about, if we are given the set {5,6,8}, we can only arrange them one way: in numerical order: 4568. We cannot arrange them any other way. Thus, we use combinations. If we CAN arrange them in any way, we use nPr (this function is right above the nCr function). For example:</p>
<p>The permutations of a, b, and c, taken two at a time, are:</p>
<p>ab, ba,
ac, ca, and
bc, cb</p>
<p>We see that we can rearrange the letters within each respective set {ab}, {ac}, and {bc}. Since there are 3 letters, and since they are taken 2 at a time, 3P2 should yield 6 (there are 6 permutations, as listed above). It indeed does.</p>
<p>The combinations of a, b, and c, taken two at a time, are:</p>
<p>ab,
ac, and
bc</p>
<p>3C2 = 3 combinations.</p>
<p>LOL you totally lost me there crazybandit.</p>
<p>dont you learn that in trigonometry/pre-calculus? im sorry, lol. just dont read it. you really really dont have to learn it for the SAT.</p>
<p>crazybandit, i agree with your solution, which is indeed the correct mathematical approach, but i wouldn’t consider it “clear-cut” though. =P</p>
<p>Crazybandit, I think listing out the possible outcomes is more simpler than the longer way of applying the combination method because it doesn’t take so long unlike the other problems that are similar to the OP’s.</p>
<p>Crazybandit–</p>
<p>Yours is a very neat solution to the problem! (And I agree, not the fastest or most straight-forward, but still…)</p>
<p>In your explanation, you say you used 5C3 instead of 5P3 because order matters. I think I know what you meant, but usually combinations are when order doesn’t matter and permutations are when it does. But THIS time, we just get to choose the combination of digits from 5,6,7,8,9 and then the problem requires that they be written in ascending order.</p>
<p>One other neat thing: while I was thinking about your solution, I tried a shorter version of the same problem: Same question, but now using #s from 400 to 500. Guess what? Same answer! List them out if you don’t believe it – but using your way, it’s just 5C2. But 5C2 = 5C3 — in general, nCr=nC(n-r) every time.</p>
<p>Anyway, thanks for your post.</p>
<p>Thanks to all!
Crazybandit-she actually really understands your explanation. I guess we all learn differently! :)</p>
<p>wow crazybandit…thanks for the tip with TI-83…I spend so much time on combo problems and I just learned the nCr thing tonight from Grubers, and my calc or the formula will save so much time. </p>
<p>Thanks!</p>
<p>learning how to do combinations/permutations is useless for the SAT unless you’re taking subject test math2. Just sayin’… The questions on the SAT, as many have noted, do not require you to bust out nCr; simple counting techniques would suffice.</p>