Question

On Halloween, a man presents a child with a bowl containing eight different pieces of candy. He tells her that she may have three pieces. How many choices does she have

Answer 1

`56` choices

Explanation:

We know that we'll have to solve this problem with a permutation or a combination, but which one do we use? The answer is a combination because the order in which the child picks the candy does not matter.

To further demonstrate this, imagine I have 4 pieces of candy labeled A, B, C, and D. I could choose A, then C, then B or I could choose C, then B, then A, but in the end, I still have the same pieces, regardless of what order I pick them in. I hope that helps to understand why this problem will be solved with a combination.

Anyways, back to the solving! Remember that the combination formula is

`_nC_r=(n!)/(r!(n-r)!)`, where n is the number of objects in the sample (the number of objects you choose from) and r is the number of objects that are to be chosen.

In this case,
`n=8` and
`r=3`. Substituting these values into the formula gives us:

`_8C_3=(8!)/(3!5!)`

`= (8*7*6*5*4*3*2*1)/(3*2*1*5*4*3*2*1)` (Expand the factorials)

`=(8*7*6)/(3*2*1)` (Cancel out
`5*4*3*2*1`)

`=(8*7*6)/(6)` (Evaluate denominator)

`=8*7` (Cancel out
`6`)

`=56`

Therefore, the child has
`\bf56` different ways to pick the candies. Hope this helps!