1
$\begingroup$

A basket holds 10 bananas, 8 apples, 6 oranges and 4 pears. Let $a_i$ be the number of different presents composed of $i$ fruits. Each such present contains at least 1 banana, at most 2 apples, the number of oranges in a present is even, and the number of pears in a present is odd. Create a generating function for ${\{{a_i\}}^\infty_{i=1}}$.

I'm not really sure how to approach this problem, I'd appreciate any guidance.

$\endgroup$
1
$\begingroup$

The generating function \begin{eqnarray*} (b+b^2+\cdots+b^{10})(1+a+a^2)(1+o^2+o^4+o^6)(p+p^3) \end{eqnarray*} will give the number of choices of fruits and keep tabs on which type of each fruit.

We are only intrested in the number of configurations for a given number of fruits, so let $a=b=o=p=x$ and we have \begin{eqnarray*} a_i= [x^i]:(x+x^2+\cdots+x^{10})(1+x+x^2)(1+x^2+x^4+x^6)(x+x^3). \end{eqnarray*}

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Shouldn't the sequance of a's start at 1 + a + ... etc and end at a^2, if there's at most 2 apples? $\endgroup$ – user747644 Feb 14 at 13:28
  • $\begingroup$ @user747644 Yes. I had misread the question. Thank you for bringing this to my attention. $\endgroup$ – Donald Splutterwit Feb 14 at 13:31
  • $\begingroup$ So the value of let's say $a_5$ is the coefficient of $x^5$ in the generating function? $\endgroup$ – user747644 Feb 14 at 13:34
  • $\begingroup$ That's right. $[x^i]:f(x)$ is a coefficient extractor ... it means the coefficient of $x^i$ in the function $f(x)$. $\endgroup$ – Donald Splutterwit Feb 14 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.