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Textbook Question:

Textbook Solution:

a) $31 \choose 12$ --> Regular combinations

b) $31+12-1\choose 12$ --> Combinations with repetition

c) $42 \choose 12$ - 31

Personal Logical Questions:

  1. Part A: Is the limiting factor (r) the number of flavors or the number of times a flavor can be ordered ? How do you know ?

  2. Part C: For getting $42 \choose 12$ , we do $31+12-1 \choose 12$ ?

  3. Part C: Why the value of r is not 11 ? Where does - 31 comes from ?

1 Answers 1

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For A, the key is that a flavor cannot be repeated - this puts us in the standard "combinations" situation - selection of 12 distinct objects (flavors) from a set of 31 - so $\binom{31}{12}$ possible selections.

For B, I believe the answer given has an error - should be $\binom{31+12-1}{12}$ (or, equivalently, $\binom{31+12-1}{30}$)- this is the standard formula for selecting 12 objects from 31 "types" (allowing repetition of types): Selecting $r$ objects from $n$ types with free repetition allowed can be done in $\binom{n+r-1}{n-1} = \binom{n+r-1}{r}$ different ways. The given result seems to have confused $r-1$ with $n-1$

For C, we start with the total number of ways to select with no restrictions (corrected result from B) and subtract number the non-allowed selections - We are not allowed to have all 12 cones of the same flavor, and there are 31 flavors to choose - so there are 31 forbidden selections; thus $\binom{31 +12-1}{12}-31$ allowed selections

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    Helpful information! I am still confused on part C, the 31 flavors. The question says "a flavor cannot be repeated more than 11 times". Is the 11 referring to the # of cones ? If there are 12 cones, and each cone can have 31 flavors, but we now are allowed to have 11 cones. Then there are 1 cone x 31 flavors less. Am I right with my logic ? – 2017-02-21