Problem involving combinations. In how many ways can $42$ candies (all the same) be distributed among 6 different infants such that each infant gets an odd number of candies?
I seem to think that we have 42 different objects, and 6 choices. So it should be 42C6. However, I'm not factoring in the "odd number of candies" part of the question, so I'm sure it's wrong. Any help is appreciated.

We will give out $2x_i+1$ candies to the $i$-th infant. Thus we want $2(x_1+\cdots+x_6)+6=42$, or equivalently $x_1+\cdots+x_6=18$.
Counting the number of solutions $(x_1,\dots,x_6)$ in non-negative integers of the equation $x_1+\cdots+x_6=18$ is a standard Stars and Bars problem. 
Or else let's do the distribution cruelly. We will distribute $48$ candies to the infants, a positive even number of candies to each infant, and then take away a candy from each infant. 
Tie the candies in bundles of $2$ each, and line up the bundles like this:
$$ \infty \quad\infty\quad\infty\quad \infty \quad\infty\quad\infty\quad\dots \quad\infty \quad\infty\quad\infty\quad\infty.$$
In the usual way, we want to choose $5$ intercandy gaps from the $23$ to put a separator into.  
