In how many ways can you write 18 as the sum of three counting numbers?
hmm what class is this for?
algebra 2/Trig
Is this a homework problem, or are you just curious? This is a question for combinatorics.
this is review for the test tomorrow. I think my teacher is trying to start us on pre calc for next year so i think thats what this is.
Do you have any idea how you are supposed to solve this?
on these type of combination problems we've been using factorials, and sometimes exponential functions, but I don't know where to start on this one
3! = 6 6 + 6 + 6 10+7+1 10+5+3 10+6+2 .......
are counting numbers just natural?
so its 3 ! ?
I can list much more than three if I understand the question right. But I don't think I do
err much more than 6
we've also been using pascals triangle if that helps.
@Luis_Rivera can you explain please
is it possible that it has something to do with the 18th row? or 18!/(3!15!) or something?
I have no idea, maybe luis will answer.
haha I hope so
but I can list way more than 6 ways of writing 18 with the sum of three natural numbers
10 + 5 + 3= 5 *2 + 5 + 3 primez : 5, 3, and 2 6 + 4 + 8 = 3*2 + 2*2 + 2^3 And on and on, notice that even though you see a 5, the actual numbers used total 3. Anyway that's how I was taught in ancient India
ahh I see.
so its how many numbers are used, not 3 numbers added together.
bah im still confused...lol gl @liv2dance
We seem to be ignoring repeats. it does NOT say 3 DISTINCT values. Try expanding this \(\left(\sum\limits_{n = 1}^{16}x^{n}\right)^{3}\), and just reading the coefficient off the resulting \(x^{18}\) term. I get 136.
I know it does not say distinct values, but they must be assumed, like in permutations of a word. Even though they do not say distinct you will not use the same word twice
i think @tkhunny is right though.
I don't see what the prime numbers have to do with finding the amount of (__+__+__) that equals 18
Generating function can be tedious, to be sure, but the idea is not overwhelming.
Haha sorry. I'm almost positive @tkhunny is right. Thanks for everyones help though
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