Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

6: 5. Combinations

  • Page ID
    25691
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    The following topics are included in this series of six videos.

    1. Intro to Combinations
    2. Combination Notation
    3. Combination or Permutation?
    4. Combinations with Cases
    5. Combinations with Stages
    6. Combinations with Complements

     

     

     

     

    The problem in the following video is solved in two ways: first using cases and stages, and then using complements.

     

    PREWORK:

    1. A student has volunteered to bring the chips for a pot-luck party. She is in a hurry and runs into a grocery store to get 3 different kinds of chips. There are 10 different kinds of chips on the shelf. How many ways can she choose the 3 bags of chips if they all have to be different kinds?
    2. Out of Sarah, Jennifer, Jason, and Quentin, I randomly select at least one person but at most three people to take the survey, how many groups could I choose?
    3. If a combination lock has 40 numbers, how many different "combinations" are there (assume a standard combination lock requires 3 numbers to unlock)? Should we really be using the word "combination"? 

    Solutions:

    1. Since the student is getting different kinds of chips, repeats are not allowed. Furthermore, the order in which she purchases the chips does not matter as they all will end up at the party. Therefore we use a combination, so the answer is \(C(10,3)\).

    2. We can use combinations here since no one will take the survey more than once (no repeats) and since all participants have the same roll: taking the survey. There are three cases to consider: exactly 1 person takes  the survey, exactly 2 people take the survey, exactly 3 people take the survey. Therefore, the answer is \(C(4,1)+C(4,2)+C(4,3)\).

    3. The numerical answer depends on how a combination lock works. First, notice that the order in which the numbers are entered on such a lock definitely matters. If all of the numbers can be repeated, there are \(40\cdot 40\cdot 40\) options. If repeats are not allowed, then the answer is \(P(40,3)\). If repeats are allowed but not in a row, then the answer is \(40\cdot 39\cdot 30\). Regardless, none of these are combinations!