Determine the grouping of specific candies to a grouping of 5, 4, 3, 2 , or 1 per basket.
NOTE= All baskets are identical.
$$5\ -\ 0\ -0\ -\ 0\ =5c5=\frac{\left(5\cdot4\cdot3\cdot2\cdot1\right)}{5!}=\frac{120}{5\cdot4\cdot3\cdot2\cdot1}=\frac{120}{120}=1$$
$$4-1-0-0=5c4\cdot1c1=\frac{\left(5\cdot4\cdot3\cdot2\right)}{4\cdot3\cdot2\cdot1}\cdot\frac{1}{1}=\frac{120}{24}=\frac{60}{12}=5$$
$$3-2-0-0=5c3\ \cdot\ 2c2=\frac{\left(5\cdot4\cdot3\right)}{3\cdot2\cdot1}\cdot\frac{\left(2\cdot1\right)}{2\cdot1}=\frac{60}{6}\cdot\frac{2}{2}=10$$
In order to account for the fact that order does not matter and also to avoid double counting we will divide the number of duplicate baskets.
$$3-1-1-0=\frac{\left(5c3\cdot2c1\cdot1c1\right)}{\left(2!\right)}$$
$$\frac{=10\ \cdot\ 2\ \cdot\ 1}{2\cdot1}=\frac{20}{2}=10$$
$$2-2-1-0=\frac{\left(5c2\cdot3c2\cdot1c1\right)}{2!}$$
$$\frac{\left(=\frac{\left(5\cdot4\right)}{2\cdot1}\cdot\frac{\left(3\cdot2\right)}{2\cdot1}\cdot\frac{\left(1\cdot1\right)}{1\cdot1}\right)}{2\cdot1}$$
$$=\frac{\left(10\cdot3\cdot1\right)}{2}=\frac{30}{2}=15$$
$$2-1-1-1=\frac{\left(5c2\cdot3\cdot2\cdot1\right)}{3!}$$
$$\frac{\left(10\cdot3\cdot2\cdot1\right)}{3\cdot2\cdot1}=10$$
Total = 1 + 5 + 10 + 10 15 + 10
=51
Option B is CORRECT.
