Find the number of ways to select $1,2,3,4,5,6,7,8,9,10$ fruits from a pile (using generating functions) Find the number of ways to select $1,2,3,4,5,6,7,8,9,10$ fruits from a pile of $3$ apples, $5$ oranges and $2$ bananas. (Use generating functions.)
Any tips? 
 A: We encode zero up to


*

*three apples as: $\ \quad1+x+x^2+x^3=\frac{1-x^4}{1-x}$

*five oranges as: $\ \quad1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x}$

*two bananas as: $\quad 1+x+x^2=\frac{1-x^3}{1-x}$

Denoting with $[x^k]$ the coefficient of $x^k$ of a series, we are looking for
  \begin{align*}
\color{blue}{[x^k]\frac{(1-x^4)(1-x^6)(1-x^3)}{(1-x)^3}}\qquad \qquad 1\leq k\leq 10
\end{align*}
We  obtain
  \begin{align*}
[x^k]&\frac{(1-x^4)(1-x^6)(1-x^3)}{(1-x)^3}\\
&=[x^k](1-x^4)(1-x^6)(1-x^3)\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\tag{1}\\
&=[x^k](1-x^3-x^4-x^6+x^7+x^9+x^{10})\sum_{j=0}^\infty\binom{j+2}{2}x^j\tag{2}\\
\end{align*}

Comment:


*

*In (1) we apply the binomial series expansion.

*In (2) we expand the factors up to powers of $x^{10}$ since higher powers do not contribute to $[x^k]$ for $1\leq k\leq 10$. We also apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

Example: We calculate from (2) the coefficient of  $x^k$  for $k=7$.
  \begin{align*}
\color{blue}{[x^7]}&\color{blue}{(1-x^3-x^4-x^6+x^7+x^9+x^{10})\sum_{j=0}^\infty\binom{j+2}{2}x^j}\\
&=\left([x^7]-[x^4]-[x^3]-[x^1]+[x^0]\right)\sum_{j=0}^\infty\binom{j+2}{2}x^j\\
&=\binom{9}{2}-\binom{6}{2}-\binom{5}{2}-\binom{3}{2}+\binom{2}{2}\\
&=36-15-10-3+1\\
&\,\,\color{blue}{=9}
\end{align*}

The  $9$  solutions  of seven fruits are
$$
\begin{array}{c|c|c}
\text{apples}&\text{oranges}&\text{bananas}\\
\hline
3&4&0\\
3&3&1\\
3&2&2\\
2&5&1\\
2&4&1\\
2&3&2\\
1&5&1\\
1&4&2\\
0&5&2\\
\end{array}
$$
A: I am guessing, you consider choosing different fruits of same variety as identical choice, otherwise question will become straightforward. 
So, suppose you have to pick $n$ fruits, where $n =1,2,3,...10$
You can seek no. of solution of equation $x_1 +x_2 +x_3 = n$,  for different values of $n$, where $x_1,x_2,x_3$ represent no. of apples, oranges and bananas. 
By, generating function, you know it is $\binom {2+n}n$ 
For different values of $n$, it becomes $\sum_{n=1}^{10}\binom {2+n}n= \sum_{n=1}^{10}\binom {2+n}2 = \sum_{n=3}^{12}\binom {n}2 = \frac12 [ \sum_{n=3}^{12} n^2- \sum_{n=3}^{12}n] =285 $
