Candies to other Children

Casework would seem kind of messy. So instead we will try to make a generating function.

The generating function for the youngest two kids are : $(1+x)^2$

because they only want zero or one candy.

The generating function for the middle kid is: $1+x+x^2+...$

because he is fine with any number of candy.

The generating function for Calvin and Lin are: $x+x^3+x^5+...)^2$

since they only want an odd number of candy.

Therefore the number of ways to distribute the 100 candies is equivalent to the coefficient of the $x^{100}$ term of:

$(1+x)^2(1+x+x^2+...)(x+x^3+x^5+...)^2 = \frac{x^2(1+x)^2}{(1-x)(1-x^2)^2}$

$= \frac{x^2}{(1-x)^3} = x^2\sum_{k=0}^{\infty}{{k+2 \choose 2}x^k}$

It is easy to see that the $x^{100}$ term coefficient is just the $x^{98}$ term coefficient of $\sum_{k=0}^{\infty}{{k+2 \choose 2}x^k}$ . This implies that the number of ways to distribute the candy is just ${100 \choose 2} = \boxed{4950}$ .

Notice how $100 \choose 2$ is the number of ways to sort $101$ identical candies into three non-empty piles. I'll leave it up to the reader to find a one to one correspondence. :)