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A person can feel the four basic flavors: sour, sweet, salty, bitter. Any different flavor is Combination of some of the flavors mentioned above.

  1. How much different falvors we have in total?
  2. How much flavors we have that are combination of 3 basic flavors?

if each flavor is combination of the others so we have $4$ basic flavors so its like $n^{n-1}$?
what about the second? I would like to get some advice how to do that
Thanks!

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1  
What is $n$ in "its like $n^{n-1}$"? –  Phira Aug 17 '13 at 13:40
    
I tought its like $4^3$ but I wrong. –  Ofir Attia Aug 17 '13 at 13:43
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How do you know you're wrong? Do you have the correct answers? If so, you should include them in the question and explain that you don't know how to get them. –  Michael Albanese Aug 17 '13 at 13:51

2 Answers 2

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HINTS:

  1. The set $B=\{\text{sour},\text{sweet},\text{salty},\text{bitter}\}$ of basic flavors has four elements. Each flavor is a combination of one or more of these four basic flavors, so each flavor corresponds to a non-empty subset of $B$. How many subsets does a $4$-element set have? How many of them are non-empty?

  2. Choosing a set of $3$ basic flavors to combine is essentially the same thing as choosing one basic flavor to leave out of the combination. How many ways are there to choose one missing basic flavor?

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its like to say for (2) $\binom {3}{4}\cdot \binom{2}{4} \cdot \binom {1}{4}$? –  Ofir Attia Aug 21 '13 at 6:47
    
@Ofir: You’re making it way too hard (and you’ve got your binomiral coefficients upside down). There are four basic flavors; to make a flavor that is a combination of three of them, you must leave one out. You can leave out any one of the four, so there are $4$ ways to pick $3$ flavors to combine. If you prefer to use binomial coefficients, there are $\binom43$ ways to pick $3$ of the $4$ flavors, and that’s the only choosing that you need to do to build a flavor that’s a combination of $3$ basic flavors. –  Brian M. Scott Aug 21 '13 at 6:49
    
for (2) I can consider it as $\mathcal {P}(B)$? –  Ofir Attia Aug 21 '13 at 6:55
    
@Ofir: I don’t understand what you mean. For (2) you just observe that there are $\binom43=4$ ways to pick $3$ of the $4$ basic flavors to use in forming a combination; that’s all there is to it. –  Brian M. Scott Aug 21 '13 at 7:00
    
Oh I wrote (2), its for (1) –  Ofir Attia Aug 21 '13 at 7:02

Hint: The number of ways of choosing $k$ flavours from $4$ is ${4 \choose k}$. For all possible flavours, consider all the possible values $k$ can have. For the second question, you only need to consider one value of $k$.

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