<p>What are all of the methods for tackling these?</p>
<p>For example:</p>
<p>Group A has 3 workers.
Group B has 2 workers.</p>
<p>How many combinations are possible with 2 workers from group A and 1 from group B?</p>
<p>Please include methods for probability.</p>
<p>I'm really not that good at these problems. So I'll try to answer just so I can practice.
Ok so there are a total of 3 spots ( right, 2 for a and 1for B)</p>
<p>so for the first two slots I can pick 3C2 and for the other slot I can pick 2C1, so (3C2)(2C1) is 6....that doesn't sound right...crap.</p>
<p>6 looks right to me... i just did it like this</p>
<p>Group A: A B C
Group B: D E</p>
<p>ABD
ABE</p>
<p>ACD
ACE</p>
<p>BCD
BCE</p>
<p>I would like to know this too, probability/combinations are my achilles heel</p>
<p>btw greenbay I guarentee you that any problem # over 12 will not be that easy</p>
<p>and the CB has a NASTY habit of keeping probability has the second to last or last question</p>
<p>yeah lol that would be a question under 15.. 15+ are the hard ones</p>
<p>But is there a definate equation/concrete method for all of these? And how about problems involving factorials (as one way to solve them)?</p>
<p>I would really like to know a better way of doing these. Writing out all the combinations is were I always make stupid mistakes. It also takes so long!</p>
<p>ok so yay, I got it right. What I 've done up to know has led me to the conclusion: there is no easy way out, you've got to reason through it.</p>
<p>I'll leave you guys with this problem
there are 5 math books, 4 physics books, and 3 bio books, if the books of the same subject must always be kept together in how many different ways can they be ordered on a shelf?</p>
<p>¨btw this one is a factorail one for the person who asked about factorial problems...reason trhough it</p>
<p>
<p>if the books of the same subject must always be kept together (meaning they would have to be touching each other) then you can take the 5 math books = 1 big math book, 4 physics book = 1 big physics book, 3 bio books = 1 big bio book.</p>
<p>so you got 3 'books' and you gotta order them on a shelf in as many different ways as possible.</p>
<p>3 nPr 3 I believe would be the function. I believe the answer is 6</p>
<p>3 nPr 3 = 3!
3! = 3x2x1 = 6</p>
<p>let's reason through it.</p>
<p>M = math book
P = physics book
B = bio book
left = top
right = bottom</p>
<p>M P B
M B P
P M B
P B M
B M P
B P M</p>
<p>so yea, 6. same answer as previous problem :X</p>
<p>ok you have part of the problem, yes the three "stacks" of subjects can be ordered in 3!...but remember that the books within the stacks can be ordered in a different manner as well. So the total number of different arrangments would be:
3! * (5!<em>4!</em>3!)</p>
<p>here's another one, this is a really cool one: There are 5 men and 8 women. Withing those 5 men and 8 women there are a total of 3 couples ( married, dating, whatever). You must form a committee of 5 with NO couples in it, how many different committees can you have?</p>
<p>here's another one, this is a really cool one: There are 5 men and 8 women. Withing those 5 men and 8 women there are a total of 3 couples ( married, dating, whatever). You must form a committee of 5 with NO couples in it, how many different committees can you have?</p>
<p>INFO:</p>
<p>5 men, 8 women, 13 people.</p>
<p>3 couples--6 people invovled in a couple, 7 uninvolved</p>
<p>Divide in groups . . . . </p>
<p>Group A: 3 involved men, 7 uninvolved OR
Group B: 3 involved women, 7 uninvolved</p>
<p>Re-distribute to three groups . . . </p>
<p>Group A: 3 involved men
Group B: 7 uninvolved
Group C: 3 involved women</p>
<p>How many groups of 5 are possible without A and Cs being in the group at once?</p>
<p>Now I'm stuck.</p>
<p>This is an easy one. Since you know that the number of singles is 7 and you need to SELECT a committee of 5. There are 7C5 ways. 21 ways. So 21 different committees.</p>
<p>I think you misread the question, how many different combinations are possible with NO COUPLES. Meaning you have to know the combinations of people possible from Group A and Group C also, without ever mixing group A and C so no couple will be in the committee.</p>
<p>Here's one possible way to answer the question:</p>
<h1>committees with NO couples =</h1>
<p>(total #possible committees) - (#committees with at least ONE couple)
= A - B (for simplicity)</p>
<p>A = 13 nCr 5 = 13x12x11x10x9 / (5x4x3x2x1) = 1287</p>
<p>To compute B, compute #combinations with exactly ONE couple + #with exactly TWO couples. (Can't have 5_person committee with three couples, would be too many people).</p>
<p>B = (3)( 7 nCr 3) + (3 nCr 2) ( 7 nCr 1)
= (3)(7 x 6 x 5)/(3x2) + (3)(7)
= 105 + 21
= 126</p>
<p>So answer = A - B = 1287 - 126 = 1161.</p>
<p>...unless someone sees a flaw in the logic or math...</p>
<p>Ha ha, nothing this hard ever shows up on the SAT. You only get basic or extended applications of the counting principle (i.e. multiply), and perhaps something that could be listed out or plugged into a TI as 15 nCr 3. Even the ones that have shown up on the Math 2 are pretty formulaic.</p>
<p>hey could somone explain the ncr thing to me?</p>
<p>optimizerdad is right, I was gonna type up the right answer today in the morning but then the school bus came by..</p>
<p>greenbay nCr denotes a combination ( think poker, where the order in which you get the cards doesn't matter but what matters are which cards you have period) where n is the number of elements and r are the spots.</p>
<p>nCr= (n!)/((r!)(n-r)!)</p>
<p>GreenBay:
nCr is the symbol for 'How many combinations are there of n items, taken r at a time'.</p>
<p>I have a statement to confirm.</p>
<p>nPr is used for specific ordering, such as 1 2 3 vs 1 3 2 vs 2 1 3 etc. </p>
<p>nCr is used for specific combinations, regardless of their order, right?</p>
<p>Also, let's say you had 30 plumbers and you wanted to know how many combinations of two were possible. Would this be 30nCr2? Meaning, would this algorithm omit the repetiion of couples?</p>