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

Answer:

$56$ choices

Step-by-step 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

$_nC_r=\frac{n!}{r!(n-r)!}$, 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:

$_8C_3=\frac{8!}{3!5!}$

$= \frac{8*7*6*5*4*3*2*1}{3*2*1*5*4*3*2*1}$ (Expand the factorials)

$=\frac{8*7*6}{3*2*1}$ (Cancel out $5*4*3*2*1$)

$=\frac{8*7*6}{6}$ (Evaluate denominator)

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

$=56$

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