{"id":1749,"date":"2025-07-28T19:01:54","date_gmt":"2025-07-28T19:01:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1749"},"modified":"2026-03-26T16:59:56","modified_gmt":"2026-03-26T16:59:56","slug":"counting-principles-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/counting-principles-learn-it-4\/","title":{"raw":"Counting Principles: Learn It 4","rendered":"Counting Principles: Learn It 4"},"content":{"raw":"

Combinations<\/h2>\r\nSo far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations<\/strong>. A selection of [latex]r[\/latex] objects from a set of [latex]n[\/latex] objects where the order does not matter can be written as [latex]C\\left(n,r\\right)[\/latex]. Just as with permutations, [latex]\\text{C}\\left(n,r\\right)[\/latex] can also be written as [latex]{}_{n}{C}_{r}[\/latex]. In this case, the general formula is as follows.\r\n

[latex]\\text{C}\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\r\n\r\n

\r\n

formula for combinations of [latex]n[\/latex] distinct objects<\/h3>\r\nGiven [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects where the order does not matter<\/strong> from the set is\r\n

[latex]\\text{C}\\left(n,r\\right) = {}_{n}{C}_{r} =\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\r\n\r\n<\/section>

An earlier problem considered choosing [latex]3[\/latex] of [latex]4[\/latex] possible paintings to hang on a wall. We found that there were [latex]24[\/latex] ways to select [latex]3[\/latex] of the [latex]4[\/latex] paintings in order. But, what if we did not care about the order? We would expect a smaller number because selecting paintings [latex]1, 2, 3[\/latex] would be the same as selecting paintings [latex]2, 3, 1[\/latex].\r\n[latex]\\\\[\/latex]\r\nTo find the number of ways to select [latex]3[\/latex] of the [latex]4[\/latex] paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order [latex]3[\/latex] paintings.\r\n[latex]\\\\[\/latex]\r\nThere are [latex]3!=3\\cdot 2\\cdot 1=6[\/latex] ways to order [latex]3[\/latex] paintings.Thus, there are [latex]\\frac{24}{6} = 4[\/latex] ways to select [latex]3[\/latex] of the [latex]4[\/latex] paintings.\r\n[latex]\\\\[\/latex]\r\nUsing the formula:\r\n

[latex]\\text{C}\\left(4,3\\right) = {}_{4}{C}_{3} =\\dfrac{4!}{3!\\left(4-3\\right)!} = \\dfrac{4 \\cdot 3 \\cdot 2 \\cdot 1}{(3 \\cdot 2 \\cdot 1)(1)} = \\dfrac{24}{6} = 4[\/latex] ways.<\/p>\r\n\r\n<\/section>

Note the similarity and difference between the formulas for permutations and combinations:\r\n
    \r\n \t
  • Permutations (order matters), [latex]P(n, r)=\\dfrac{n!}{(n-r)!}[\/latex]<\/li>\r\n \t
  • Combinations (order does not matter), [latex]C(n, r)=\\dfrac{n!}{r!(n-r)!}[\/latex]<\/li>\r\n<\/ul>\r\nThe formula for combinations is the formula for permutations with the number of ways to order [latex]r[\/latex] objects divided away from the result.\r\n\r\n<\/section>
    A fast food restaurant offers five side dish options. Your meal comes with two side dishes.\r\n
      \r\n \t
    1. How many ways can you select your side dishes?<\/li>\r\n \t
    2. How many ways can you select [latex]3[\/latex] side dishes?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"218844\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"218844\"]\r\n
        \r\n \t
      1. We want to choose [latex]2[\/latex] side dishes from [latex]5[\/latex] options.\r\n[latex]\\text{C}\\left(5,2\\right)=\\dfrac{5!}{2!\\left(5 - 2\\right)!}=10[\/latex]<\/li>\r\n \t
      2. We want to choose [latex]3[\/latex] side dishes from 5 options.\r\n[latex]\\text{C}\\left(5,3\\right)=\\dfrac{5!}{3!\\left(5 - 3\\right)!}=10[\/latex]<\/li>\r\n<\/ol>\r\nUsing Calculator:\r\n[latex]\\\\[\/latex]\r\n<\/strong>We can also use a graphing calculator to find combinations. Enter [latex]5[\/latex], then press [latex]{}_{n}{C}_{r}[\/latex], enter [latex]3[\/latex], and then press the equal sign. The [latex]{}_{n}{C}_{r}[\/latex], function may be located under the MATH menu with probability commands.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
        [ohm_question hide_question_numbers=1]321994[\/ohm_question]<\/section>
        [ohm_question hide_question_numbers=1]321995[\/ohm_question]<\/section>
        [ohm_question hide_question_numbers=1]321996[\/ohm_question]<\/section>","rendered":"

        Combinations<\/h2>\n

        So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations<\/strong>. A selection of [latex]r[\/latex] objects from a set of [latex]n[\/latex] objects where the order does not matter can be written as [latex]C\\left(n,r\\right)[\/latex]. Just as with permutations, [latex]\\text{C}\\left(n,r\\right)[\/latex] can also be written as [latex]{}_{n}{C}_{r}[\/latex]. In this case, the general formula is as follows.<\/p>\n

        [latex]\\text{C}\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n

        \n

        formula for combinations of [latex]n[\/latex] distinct objects<\/h3>\n

        Given [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects where the order does not matter<\/strong> from the set is<\/p>\n

        [latex]\\text{C}\\left(n,r\\right) = {}_{n}{C}_{r} =\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n<\/section>\n

        An earlier problem considered choosing [latex]3[\/latex] of [latex]4[\/latex] possible paintings to hang on a wall. We found that there were [latex]24[\/latex] ways to select [latex]3[\/latex] of the [latex]4[\/latex] paintings in order. But, what if we did not care about the order? We would expect a smaller number because selecting paintings [latex]1, 2, 3[\/latex] would be the same as selecting paintings [latex]2, 3, 1[\/latex].
        \n[latex]\\\\[\/latex]
        \nTo find the number of ways to select [latex]3[\/latex] of the [latex]4[\/latex] paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order [latex]3[\/latex] paintings.
        \n[latex]\\\\[\/latex]
        \nThere are [latex]3!=3\\cdot 2\\cdot 1=6[\/latex] ways to order [latex]3[\/latex] paintings.Thus, there are [latex]\\frac{24}{6} = 4[\/latex] ways to select [latex]3[\/latex] of the [latex]4[\/latex] paintings.
        \n[latex]\\\\[\/latex]
        \nUsing the formula:<\/p>\n

        [latex]\\text{C}\\left(4,3\\right) = {}_{4}{C}_{3} =\\dfrac{4!}{3!\\left(4-3\\right)!} = \\dfrac{4 \\cdot 3 \\cdot 2 \\cdot 1}{(3 \\cdot 2 \\cdot 1)(1)} = \\dfrac{24}{6} = 4[\/latex] ways.<\/p>\n<\/section>\n

        Note the similarity and difference between the formulas for permutations and combinations:<\/p>\n
          \n
        • Permutations (order matters), [latex]P(n, r)=\\dfrac{n!}{(n-r)!}[\/latex]<\/li>\n
        • Combinations (order does not matter), [latex]C(n, r)=\\dfrac{n!}{r!(n-r)!}[\/latex]<\/li>\n<\/ul>\n

          The formula for combinations is the formula for permutations with the number of ways to order [latex]r[\/latex] objects divided away from the result.<\/p>\n<\/section>\n

          A fast food restaurant offers five side dish options. Your meal comes with two side dishes.<\/p>\n
            \n
          1. How many ways can you select your side dishes?<\/li>\n
          2. How many ways can you select [latex]3[\/latex] side dishes?<\/li>\n<\/ol>\n