Answer:
[tex]56[/tex] 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
[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex], 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, [tex]n=8[/tex] and [tex]r=3[/tex]. Substituting these values into the formula gives us:
[tex]_8C_3=\frac{8!}{3!5!}[/tex]
[tex]= \frac{8*7*6*5*4*3*2*1}{3*2*1*5*4*3*2*1}[/tex] (Expand the factorials)
[tex]=\frac{8*7*6}{3*2*1}[/tex] (Cancel out [tex]5*4*3*2*1[/tex])
[tex]=\frac{8*7*6}{6}[/tex] (Evaluate denominator)
[tex]=8*7[/tex] (Cancel out [tex]6[/tex])
[tex]=56[/tex]
Therefore, the child has [tex]\bf56[/tex] different ways to pick the candies. Hope this helps!
