{"id":2097,"date":"2015-11-12T18:30:42","date_gmt":"2015-11-12T18:30:42","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=2097"},"modified":"2015-11-12T18:30:42","modified_gmt":"2015-11-12T18:30:42","slug":"find-the-number-of-combinations-using-the-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/find-the-number-of-combinations-using-the-formula\/","title":{"raw":"Find the Number of Combinations Using the Formula","rendered":"Find the Number of Combinations Using the Formula"},"content":{"raw":"

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.\n<\/p>

[latex]\\text{C}\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\nAn earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are [latex]3!=3\\cdot 2\\cdot 1=6[\/latex] ways to order 3 paintings. There are [latex]\\frac{24}{6}[\/latex], or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are not<\/em> selecting 1 painting. There are 4 paintings we could choose not<\/em> to select, so there are 4 ways to select 3 of the 4 paintings.\n
\n

A General Note: Formula for Combinations of n<\/em> Distinct Objects<\/h3>\nGiven [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set is\n
[latex]\\text{C}\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\n<\/div>\n
\n

How To: Given a number of options, determine the possible number of combinations.<\/h3>\n
  1. Identify [latex]n[\/latex] from the given information.<\/li>\n\t
  2. Identify [latex]r[\/latex] from the given information.<\/li>\n\t
  3. Replace [latex]n[\/latex] and [latex]r[\/latex] in the formula with the given values.<\/li>\n\t
  4. Evaluate.<\/li>\n<\/ol><\/div>\n
    \n

    Example 4: Finding the Number of Combinations Using the Formula<\/h3>\nA fast food restaurant offers five side dish options. Your meal comes with two side dishes.\n
    1. How many ways can you select your side dishes?<\/li>\n\t
    2. How many ways can you select 3 side dishes?<\/li>\n<\/ol><\/div>\n
      \n

      Solution<\/h3>\n
      1. We want to choose 2 side dishes from 5 options.\n
        [latex]\\text{C}\\left(5,2\\right)=\\frac{5!}{2!\\left(5 - 2\\right)!}=10[\/latex]<\/div><\/li>\n\t
      2. We want to choose 3 side dishes from 5 options.\n
        [latex]\\text{C}\\left(5,3\\right)=\\frac{5!}{3!\\left(5 - 3\\right)!}=10[\/latex]<\/div><\/li>\n<\/ol><\/div>\n
        \n

        Analysis of the Solution<\/h3>\nWe can also use a graphing calculator to find combinations. Enter 5, then press [latex]{}_{n}{C}_{r}[\/latex], enter 3, and then press the equal sign. The [latex]{}_{n}{C}_{r}[\/latex], function may be located under the MATH menu with probability commands.\n\n<\/div>\n
        \n

        Q & A<\/h3>\n

        Is it a coincidence that parts (a) and (b) in Example 4\u00a0have the same answers?<\/h3>\nNo. When we choose r objects from n objects, we are not<\/strong> choosing [latex]\\left(n-r\\right)[\/latex] objects. Therefore, [latex]C\\left(n,r\\right)=C\\left(n,n-r\\right)[\/latex]. <\/em>\n\n<\/div>\n
        \n

        Try It 8<\/h3>\nAn ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?\n\nSolution<\/a>\n\n<\/div>","rendered":"

        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.\n<\/p>\n

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

        An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are [latex]3!=3\\cdot 2\\cdot 1=6[\/latex] ways to order 3 paintings. There are [latex]\\frac{24}{6}[\/latex], or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are not<\/em> selecting 1 painting. There are 4 paintings we could choose not<\/em> to select, so there are 4 ways to select 3 of the 4 paintings.<\/p>\n

        \n

        A General Note: Formula for Combinations of n<\/em> Distinct Objects<\/h3>\n

        Given [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set is<\/p>\n

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

        How To: Given a number of options, determine the possible number of combinations.<\/h3>\n
          \n
        1. Identify [latex]n[\/latex] from the given information.<\/li>\n
        2. Identify [latex]r[\/latex] from the given information.<\/li>\n
        3. Replace [latex]n[\/latex] and [latex]r[\/latex] in the formula with the given values.<\/li>\n
        4. Evaluate.<\/li>\n<\/ol>\n<\/div>\n
          \n

          Example 4: Finding the Number of Combinations Using the Formula<\/h3>\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 3 side dishes?<\/li>\n<\/ol>\n<\/div>\n
            \n

            Solution<\/h3>\n
              \n
            1. We want to choose 2 side dishes from 5 options.\n
              [latex]\\text{C}\\left(5,2\\right)=\\frac{5!}{2!\\left(5 - 2\\right)!}=10[\/latex]<\/div>\n<\/li>\n
            2. We want to choose 3 side dishes from 5 options.\n
              [latex]\\text{C}\\left(5,3\\right)=\\frac{5!}{3!\\left(5 - 3\\right)!}=10[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n
              \n

              Analysis of the Solution<\/h3>\n

              We can also use a graphing calculator to find combinations. Enter 5, then press [latex]{}_{n}{C}_{r}[\/latex], enter 3, and then press the equal sign. The [latex]{}_{n}{C}_{r}[\/latex], function may be located under the MATH menu with probability commands.<\/p>\n<\/div>\n

              \n

              Q & A<\/h3>\n

              Is it a coincidence that parts (a) and (b) in Example 4\u00a0have the same answers?<\/h3>\n

              No. When we choose r objects from n objects, we are not<\/strong> choosing [latex]\\left(n-r\\right)[\/latex] objects. Therefore, [latex]C\\left(n,r\\right)=C\\left(n,n-r\\right)[\/latex]. <\/em><\/p>\n<\/div>\n

              \n

              Try It 8<\/h3>\n

              An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?<\/p>\n

              Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

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