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.
 A: You are looking for compositions of $42$ into six odd parts.  If you give each child one candy and put the rest in pairs, this will be the same as the weak compositions of 18 into six parts, which is given by ${23 \choose 5}=33649$.  To prove the formula, put $24$ (pairs) in a row, then you select five places to split the row and remove one pair from each part.  This allows for not giving any more candies to one or more infants.
A: Each infant must get at least one candle, so you have 36 candles to give out in pairs to 6 infants.  Hence you have 18 pairs to give out to 6 infants.  Can you take it from there?
A: 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.  
