<p>Yes it does depend on the value of the variables. The answer will be in terms of j and k.</p>
<p>Dr. Steve, i meant that different variables will have different answers…</p>
<p>I’m not exactly sure what you mean by that. Let me give a simple example:</p>
<p>If j=2 and k=3, then the question becomes “in how many ways can 3 be written as a sum of 2 positive integers?”</p>
<p>Well let’s see: 3=1+2=2+1. So there are 2 ways.</p>
<p>Let’s do one more: j=2, k=4</p>
<p>4=1+3=2+2=3+1. So there are 3 ways.</p>
<p>So different choices of j and k will lead to a different answer. In other words, the answer to this question depends on j and k. Another way to say this is that the answer is a function of j and k. For example, maybe the answer is j^2+3k (it’s not). I gave you a hint. The answer can be expressed as either a simple permutation or combination with the j and k as part of the indices. </p>
<p>As another hint, try more specific examples until you start to see the pattern. Then ideally, by looking at your examples try to see how this is a counting problem. What exactly are you counting?</p>
<p>@Dr Steve </p>
<p>Your example is more than helpful – it is critical! Without it, I would not have known that we were to count 1+2 and 2+1 as distinct. It’s still a hard problem, but now maybe I’m on to it :)</p>
<p>Also, no one has posted an answer to my earlier counting problem:</p>
<p>How many two-letter codes can be made using only the first 8 letters of the alphabet if letters cannot be repeated and vowels cannot follow consonants? </p>
<p>I do think that question falls within SAT difficulty…</p>
<p>I’ll IM you if I figure out your puzzle before the usual CC top guns post the answer.</p>
<p>pckeller - that’s easy: 2P2 x 6P6 or 2!6!, because the vowels (a, e) have to come before to consonants (bcdfgh) of 1440</p>
<p>@pckeller</p>
<p>I always work under the assumption that writing things in different orders gives distinct objects unless specified otherwise. For example, if someone were to tell me that prime factorizations are unique, I would say “no they’re not 6=2<em>3=3</em>2. Prime factorizations are only unique up to order!” That said, I do believe you’re right. Without showing an example, the problem may be unclear, especially since addition is commutative and all. </p>
<p>That said, I think if you write out 2 or 3 more examples you’ll be able to see what the formula is. Note that there are (at least) 2 ways to recognize this formula - the first is simply by pattern recognition (after looking at enough examples), and the second is by actually thinking about what you’re counting (again, I think some examples might be necessary to realize this). Only the second of the 2 gives a deductive argument that the formula is actually correct (although I’m sure many other proofs will work - mathematical induction comes to mind).</p>
<p>@CTScoutmom --</p>
<p>Alas, not so fast! You answered for how many 8-letter codes with the vowels in front. But I’m asking for 2-letter codes where the letters are taken from the first 8 of the alphabet, no repeating allowed and vowels not allowed to come after consonants.</p>
<p>So AE is allowed. AB is allowed. BA is not allowed.</p>
<p>^ is it 44?</p>
<p>44 it is!</p>
<p>And there are two ways to go about it (besides listing).</p>
<ol>
<li><p>Break it into two cases: consonant consonant and vowel anything</p></li>
<li><p>First figure out how many two letter codes there are of any variety. Then, subtract the number of consonant vowel combos.</p></li>
</ol>
<p>p.s. I think I have Dr Steve’s answer! If I am right, it is a very simple and elegant answer. What I don’t have yet is a simple and elegant explantion to go with it! Stay tuned.</p>
<p>Is the answer for the plumber one 8?</p>
<p>Ohhhhh okay i got it now.
Could anyone solve this for me?</p>
<p>17.If p. r. and s are three different prime numbers greater than 2, and n = p * r * s, how many positive factors, including 1 and n. does n have?</p>
<p>I assumed that its 5.
I dont have the right answer though.
It was from a random thread on internet but this question ad appeared on some old previous real tests like starting from 2003 or so.</p>
<ol>
<li>If the sum of the consecutive integers</li>
</ol>
<p>from</p>
<p>-22 to x, inclusive, is 72, what is the value of x?</p>
<p>(A) 23</p>
<p>(B) 25</p>
<p>(C) 50</p>
<p>(D) 75</p>
<p>(E) 94</p>
<p>@farouz all of the integers from -22 to 22 cancel. Then count up from 23 until you reach 72.
23
+24
47
+25
72.
Therefore, x is 25. Choice B is correct.</p>
<p>@pckeller,
i just found all the possibilties then subtracted those that can’t be from those…</p>
<p>@drsteve,
By the way, i did do the taking for different variables thing (before you said the hint actually) and i did see the pattern and all that…
and that’s why I asked you in the first place if u were sure they can be expressed by j and k… because i can’t put the pattern into a nPr or nCr expression…
I guess what i couldn’t do from the beginning was<br>
“and the second is by actually thinking about what you’re counting (again, I think some examples might be necessary to realize this)” </p>
<p>the 2nd part…</p>