AlanH
Reputation
1,284
Top tag
Next privilege 2,000 Rep.
Edit questions and answers
14 34
Impact
~49k people reached

• 6 helpful flags
• 305 votes cast

 Feb 9 comment Combinatorial proof involving factorials Would it have been possible to use the identity if you hadn't changed the index of $k$? I'm not sure I would be able to spot something so subtle. Feb 5 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...? Feb 4 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? Feb 4 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. Feb 3 comment Integer solutions @BrianM.Scott Does having it "less than" vs. "less than or equal to make a difference"? Feb 3 comment Integer solutions Without the condition, it then becomes a weak composition, correct? Feb 3 comment No. of ways to arrange letters of a word (with repetition) How do you obtain $30$? Feb 1 comment How many ways are there for 10 people to have five simultaneous telephone conversations? Yeah that explains it. Thanks for the intuitive explanation. Feb 1 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? Jan 28 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'"? Jan 23 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! Jan 23 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. Jan 16 comment General counting problem. Not sure how to proceed. Could you explain why you permute both the evenings and the triples of guests? Jan 15 comment General counting problem. Not sure how to proceed. @AlexanderGruber: Thanks. I wasn't sure how to search for this though. Jan 15 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. Jan 12 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. Jan 12 comment How to approach this problem of combinatorics Nevermind! I was thinking for the general case. Jan 12 comment How to approach this problem of combinatorics Shouldn't it be 144? I believe you're missing a 2. Jan 11 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? Jan 6 comment Proof involving Stirling numbers of the second kind @BrianM.Scott Maybe it's assumed for all positive integers greater than equal to $k$?