An ice cream parlor has n toppings.

(a) How many i-topping ice cream cones are possible, where i = 0, . . . , n?

(b) Considering that an ice cream cone can have anywhere from 0 to n toppings, derive a formula for the total number of ice cream cones possible.

(c) We can also count the number of possible ice cream cones by considering each topping to be an experiment with two outcomes: on or off. Use the GBPC to determine the number of ice cream cones possible in this way.

(d) Equate your answers from parts (b) and (c) to derive an interesting identity involving binomial coefficients. (This identity can also be derived by setting x = y = 1 in the binomial theorem and has a neat implication for the sums of rows of Pascal’s triangle.)

I'm confused on the entirety of this problem, anything helps! Thank you!

For (a) I need a formula for how many possible i-toppings ice cream cones there are. We have to assume we can't double up on toppings. I have $\frac{n!}{2x}$. I don't think this is correct.

For (b) Would it be i=x, $x \in \mathbb{Z}$. Therefore the outcomes goes from 0 to n?

For (c) and (d) I have no attempt.

  • $\begingroup$ Hint for (d): $(1+1)^n = 2^n$. This is a combinatorics question rather than a probability question $\endgroup$ – Henry Sep 14 '20 at 13:19
  • $\begingroup$ Your "attempt" for (a) has an $x$ in the answer... but $x$ does not appear in the problem. Perhaps you meant $i$ instead? Even then, this will be wrong. Consider what a binomial coefficient might be useful for here. $\endgroup$ – JMoravitz Sep 14 '20 at 13:22
  • $\begingroup$ Does the basic formula work, $\frac{n!}{k!(n-k)!}$? $\endgroup$ – Wng427 Sep 14 '20 at 13:28
  • $\begingroup$ $k$ doesn't appear in the question either $\endgroup$ – JMoravitz Sep 14 '20 at 13:29
  • $\begingroup$ $\frac{n!}{i!(n-i)!}$ $\endgroup$ – Wng427 Sep 14 '20 at 13:31

(a) There are $\binom{n}{i}$ ways to select a subset of $i$ of the toppings to use, seen directly by the definition of what a binomial coefficient represents.

(b) Summing over all possible amounts of toppings that we could have selected, there are $\sum\limits_{i=0}^n \binom{n}{i}$ total ways in which we could have selected some subset of toppings regardless of size

(c) Utilizing rule of product, noting that for each topping we either use or don't use the topping, we see that there are $2\times 2\times 2\times \cdots \times 2 = 2^n$ total ways in which we could have selected some subset of toppings regardless of size

(d) We recognize that (b) and (c) both answer the same question and so their answers must be equal even if they appear different, thus we learn that $\sum\limits_{i=0}^n \binom{n}{i} = 2^n$


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