Any Math FREEKS wanna try these?

<p>Hey! Do you like math for the sake of math? Well I got some ‘fun’ questions I tried and failed to get. Can someone help me out please? These are Data Management problems so try to avoid CALC or ALGEO if possible but I do not know the answer or how to get it so I am not certain if it would involve CALC or ALGEO. Oh, please give full solutions not just an answer so I can understand your way of thinking and how to get the answer. Any help is GREATLY appreciated. Thanks.</p>

<p>Questions:
1) How many non-negative (that means zero counts) integer solutions are there to the equation
2x1 + 2x2 + x3 + x4 = 12? <em>Where all the x’s are distinct.</em>
PLEASE NOTE: the number immediately following the x is subscripted to indicate it is a different x from all other x’s. It DOES NOT mean (for example 2x1) 2 times x times 1 but simply 2 times x and the 1 just means the x is seperate from all the other x's. </p>

<p>2) How many arrangements are there of seven a’s, eight b’s, three c’s, and six d’s with no occurrence of the consecutive pairs ca or cc?</p>

<p>3) How many subsets of three different integers between 1 and 90 inclusive are there whose sum is divisible by 3?</p>

<p>4) Given a collection of 2n objects, n identical and the other n all distinct, how many different sub collections of n objects are there?</p>

<p>If someone needs clarification, please ask.</p>

<p>For number one use 'Des Carte's Rule of Signs'. Its probably better off you looking at your precalc book or the website than having me explain it to you. It should be fairly straight forward although I still fail to prove it. (however, I'm not too clear about your signs "2x1", "2x2" "x3" and "x4". Does this mean 2x, 4x, 3x, and 4x?)</p>

<p>For number two try looking at combinatrics. Its about finding the number of arrangements you can have. Its been like 5 months since I've last done combinatrics; sorry. I think I'll go review right now.</p>

<p>the answer is 23.....just 23</p>

<p>hmm.. I can only edit the text after 20 min after the post.</p>

<p>For number two try looking at combinatrics. Its about finding the number of arrangements you can have. I'm very bad at explaining things so again, it will be better to look at the textbook explanation. And a little hint is look at 'ca' and 'cc' as one letter. The answer I believe is 22!</p>

<p>I'm not too sure whether my solution to number three is correct (I'm usually wrong when I get these feelings) but I'll give it a try. All integers can be expressed in 3k+r where r=0,1,2 There are three numbers in a set and for those three numbers to be divisible by 3, all three of them has to add up such that the sum can be expressed in 3(a+b+c). </p>

<p>This leads to two possible sets:
1. Each of the three numbers in the set are multiples of 3 (i.e. they are 3a, 3b, and 3c)
2. The remainder needs to add up to a multiple of 3 (i.e. the numbers are 3a, 3b+1, and 3c+2).</p>

<p>Again doing some combinatrics for the first possible set, the number of all the possible sets can be calculated through 30 C 3</p>

<p>for the second one, its simply 30^3. (I can't fluently explain all this; it'll just make a mess and besides, I think I'm wrong!)</p>

<p>Adding these up I got the answer 31060.</p>

<p>Number 4, the answer is 2^(n+1) There is this thing where in a collection of k elements, the number of sub collection is 2^k. Since you have n+1 number of distinct elements, you obtain 2^(n+1)</p>

<p>I really am not sure about any of these questions. I wouldnt trust on it very much :P</p>

<p>To tlqkf2002:
For question one, 2x1 is simply 2x: 2x2 is 2x: x3 is just x: x4 is x. So its basically 2x + 2x + x + x = 12. BUT the x's are all distinct numbers which can be repeated. For example, the first x can be 2, second x=3, third x=1 forth x=1 which is 2(2) + 2(3) + 1 + 1= 12</p>

<p>here are 2 puzzles my math teacher gave in class:
1. An ant travels 2 miles north 1 mile east .5 miles south, .25 miles west, .125 miles north etc. How far is the point that the ant converges on from the starting point. Hints: Use infinite sequence formula, and separate the north south movements from the east west ones. I used coordinate pairs to solve it. Answer: sqrt(3.2) or 4sqrt(5)/5</p>

<ol>
<li>There are 23 nails on a board that form a 23 sided regular polygon of radius 8. What is the length of the string, if you strung up all the sides and possible diagonals? Trig involved.
My answer is 2690. I used a sigma equation. Not sure about this answer.</li>
</ol>

<p>So Mathfreek101, did you get a good score on your homework?</p>

<p>Actually gigante these questions were not my homework. They are in my review questions for an upcoming test. My teacher had hinted they might be on the test but he has a strict no helping policy for helping with the review questions because he uses similar questions on the test.</p>

<p>anyone else help? or clarify and/or elaborate on tlqkf2002's answer?</p>