A Permutations/Combinations Question and Inquiry on Good Source for Studying The Concept Lets say a burger joint offers options for customizing burgers. There are 3 types of meats and 7 condiments. A burger must include meat but may include as many or as few condiments as the customer wants. How many different burgers are possible?
Now I have been told that the answer to this question is 3(2^7). The reasoning being that we have 8 slots. The first slot for meat has 3 options. The remaining seven slots for condiments have 2 options (yes or no).
The way I was thinking about the problem was in terms of scenarios:
1st Scenario: We select one of the 3 meats and no condiments - No. of Burgers = 3
2nd Scenario: We select one of the 3 meats and 1 of the seven condiments - No. of Burgers = 3*7 (i.e. for each meat we can select any of the second condiments).
3rd Scenario: We select one of the 3 meats, then 1 of the seven condiments and then another of the remaining 6 condiments - No. of Burgers = 3*7*6.
So we will have a total of 8 scenarios if we continue and the answer should be the sum of all the no. of burgers from each scenario and would look like:
3+(3*7)+(3*7*6)+(3*7*6*5)+(3*7*6*5*4)+(3*7*6*5*4*3)+(3*7*6*5*4*3*2)+(3*7*6*5*4*3*2*1).
Why is this line of reasoning wrong?
Also, I feel as though my intuition about this subject is quite weak. Is there a free online guide/book that can help me with this topic?
 A: Your reasoning is mostly correct. You have to divide out by $i!$ to account for the fact that the order in which you choose the $i$ condiments does not matter. Observe that you are counting:
$$\binom{3}{1} \cdot \sum_{i=0}^{7} \binom{7}{i} = 3 \cdot \sum_{i=0}^{7} \binom{7}{i}$$
Note that $\sum_{i=0}^{n} \binom{n}{i} = 2^{n}$. I offer a combinatorial proof of this. Consider the set of binary strings of length $n$. There are $2^{n}$ such binary strings. $\binom{n}{i}$ counts the number of binary strings with exactly $i$ $1$'s. So the set of strings counted by $\binom{n}{i}$ and $\binom{n}{k}$ are disjoint if $i \neq k$. Thus, by rule of sum, adding up over all such binomial coefficients counts all the binary strings:
$$\sum_{i=0}^{n} \binom{n}{i} = 2^{n}$$
So:
$$3 \cdot \sum_{i=0}^{7} \binom{7}{i} = 3 \cdot 2^{7}$$
If you are looking for an additional textbook, I would suggest Bijective Combinatorics by Nicholas Loehr. It's not free, but it is easily the best textbook I've ever read. It is very thorough and rigorous, but at the same time very detailed, explanatory, and intuitive. Loehr starts at the beginning with basic counting principles, includes tons of examples, and has tons of exercises. 
