AlanH
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 Feb5 comment Proving $\text{P}(E_1 \cap E_2 | S) = \text{P}(E_1 | S) \cdot \text{P}(E_2 | S)$ if $E_1$ and $E_2$ are...? Feb4 comment Counting method for arrangements/selections The trick you use to address the end slots, why does that work? And if you pretend you have $41$ cards, why do you permute $39$ cards instead? Feb4 comment Number of integer solutions to a system of equations So does my reasoning not work? I imagined a tree, where 1 solution to the second equation results in $n$-many solutions in the first equation; hence why I took the product. Feb3 comment Integer solutions @BrianM.Scott Does having it "less than" vs. "less than or equal to make a difference"? Feb3 comment Integer solutions Without the condition, it then becomes a weak composition, correct? Feb3 comment No. of ways to arrange letters of a word (with repetition) How do you obtain $30$? Feb1 comment How many ways are there for 10 people to have five simultaneous telephone conversations? Yeah that explains it. Thanks for the intuitive explanation. Feb1 comment How many ways are there for 10 people to have five simultaneous telephone conversations? This is so much more intuitive than the textbook solution: $(10!/(2!^5))/5!$ What's the reasoning for how they got this expression? Jan28 comment LCM. What am I missing? Why is it that you chose not to make use of lcm? Also, what do you mean by "must 'have'"? Jan23 comment Combinatorics question dealing with selection So just to make sure I understand this correctly. Say $r=5$ and $s=3$. You pick the $8$ from the $50$, and then you take the five shortest from the $8$. Thanks for your help! Jan23 comment Combinatorics question dealing with selection But doing what you say in the first paragraph doesn't make sense to me because it seems like you're cheating almost. Instead of first picking $r$ people and then $s$ as the problem states explicitly, you pick $r+s$ people and then it's like you say, "Oh, well why don't I just pick the the $r$ shortest from the $r+s$ and pretend I picked them first." When in reality you didn't actually pick those $r$ people first, you picked the $r+s$ people all together. Jan16 comment General counting problem. Not sure how to proceed. Could you explain why you permute both the evenings and the triples of guests? Jan15 comment General counting problem. Not sure how to proceed. @AlexanderGruber: Thanks. I wasn't sure how to search for this though. Jan15 comment General counting problem. Not sure how to proceed. Could you solve this just using basic combinatorics? I understand the solution, but I'd prefer to stick to combinatoric methods as I'm still new to the subject. Jan12 comment I've come up with two solutions to this problem, and I don't know which is correct. Bah! How on earth did I miss that. Thanks. Jan12 comment How to approach this problem of combinatorics Nevermind! I was thinking for the general case. Jan12 comment How to approach this problem of combinatorics Shouldn't it be 144? I believe you're missing a 2. Jan11 comment How to approach this problem of combinatorics So I shouldn't even be caring about what the actual products are? If I'm reading what you wrote correctly, this is almost like how many combinations I can make of 1 apple, 2 oranges, 2 pears, 1 kiwi, and 3 mangoes? Jan6 comment Proof involving Stirling numbers of the second kind @BrianM.Scott Maybe it's assumed for all positive integers greater than equal to $k$? Jan5 comment Proof involving Stirling numbers of the second kind @BrianM.Scott Yeah, I'm looking at it right now and it says, in italics, all positive integers x.