A deck of $ 32$ cards has $ 2$ different jokers each of which is numbered $ 0$. There are $ 10$ red cards numbered $ 1$ through $ 10$ and similarly for blue and green cards. One chooses a number of cards from the deck. If a card in hand is numbered $ k$, then the value of the card is $ 2^k$, and the value of the hand is sum of the values of the cards in hand. Determine the number of hands having the value $ 2004$.
Problem
Source: CGMO 2004 P4
Tags: function, combinatorics unsolved, combinatorics
feliz
29.12.2008 14:46
The red cards are determinated by the (even) sum of their values, since an even number can be written uniquely as a sum of power of $ 2$. So the blue cards and so the green cards. The jokers have value $ 1$. So we can use none of the jokers or both ($ 2004$ is even). In the first case, we have $ S_r + S_b + S_g = 2004$, where $ S_r$, $ S_b$ and $ S_g$ are the sum of values of red cards, blue cards and green cards. Each solution gives a hand and viceversa. Then we are looking for the number of integer solutions of $ x + y + z = 1002$ (again, note that each sum is even). This is $ \binom{1004}{2}$. In the other case we get $ \binom{1003}{2}$. Our total is thus $ \binom{1004}{2} + \binom{1003}{2}$.
shadow10
22.07.2014 15:04
Let's finish it with generating function.The generating function will be
\begin{align*} G(x) &=(1+x)^2.(1+x^2)^3.(1+x^4)^3 \cdots (1+x^{1024})^3 \\ & = \frac{(1+x)^2.(1-x^{2048})^3}{(1-x^2)^3}\\ & =(1+2x+x^2-3x^{2048}-6x^{2049}-3x^{2050}+ \cdots )(1-x^2)^{-3}\end{align*}
We have to find the coefficient of $x^{2004}$ is $G(x)$.It's easy to observe $(1-x^2)^{-3}$ does not contain any odd power of $x$.So we get, $x^{2004}$ by multiplying $1$ with the $x^{2004}$ in the second factor and $x^2$ with the $x^{2002}$ in the second factor.Therefore the ans will be $ \binom{1004}{2} + \binom{1003}{2}$