{"id":2105,"date":"2015-11-12T18:30:41","date_gmt":"2015-11-12T18:30:41","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=2105"},"modified":"2015-11-12T18:30:41","modified_gmt":"2015-11-12T18:30:41","slug":"identifying-binomial-coefficients","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/identifying-binomial-coefficients\/","title":{"raw":"Identifying Binomial Coefficients","rendered":"Identifying Binomial Coefficients"},"content":{"raw":"

In Counting Principles<\/a>, we studied combinations<\/strong>. In the shortcut to finding [latex]{\\left(x+y\\right)}^{n}[\/latex], we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)[\/latex] instead of [latex]C\\left(n,r\\right)[\/latex], but it can be calculated in the same way. So\n<\/p>

[latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\nThe combination [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)[\/latex] is called a binomial coefficient<\/strong>. An example of a binomial coefficient is [latex]\\left(\\begin{array}{c}5\\\\ 2\\end{array}\\right)=C\\left(5,2\\right)=10[\/latex].\n
\n

A General Note: Binomial Coefficients<\/h3>\nIf [latex]n[\/latex] and [latex]r[\/latex] are integers greater than or equal to 0 with [latex]n\\ge r[\/latex], then the binomial coefficient<\/strong> is\n
[latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\n<\/div>\n
\n

Q & A<\/h3>\n

Is a binomial coefficient always a whole number?<\/h3>\nYes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number. <\/em>\n\n<\/div>\n
\n

Example 1: Finding Binomial Coefficients<\/h3>\nFind each binomial coefficient.\n
  1. [latex]\\left(\\begin{array}{c}5\\\\ 3\\end{array}\\right)[\/latex]<\/li>\n\t
  2. [latex]\\left(\\begin{array}{c}9\\\\ 2\\end{array}\\right)[\/latex]<\/li>\n\t
  3. [latex]\\left(\\begin{array}{c}9\\\\ 7\\end{array}\\right)[\/latex]<\/li>\n<\/ol><\/div>\n
    \n

    Solution<\/h3>\nUse the formula to calculate each binomial coefficient. You can also use the [latex]{n}_{}{C}_{r}[\/latex] function on your calculator.\n
    [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\n
    1. [latex]\\left(\\begin{array}{c}5\\\\ 3\\end{array}\\right)=\\frac{5!}{3!\\left(5 - 3\\right)!}=\\frac{5\\cdot 4\\cdot 3!}{3!2!}=10[\/latex]<\/li>\n\t
    2. [latex]\\left(\\begin{array}{c}9\\\\ 2\\end{array}\\right)=\\frac{9!}{2!\\left(9 - 2\\right)!}=\\frac{9\\cdot 8\\cdot 7!}{2!7!}=36[\/latex]<\/li>\n\t
    3. [latex]\\left(\\begin{array}{c}9\\\\ 7\\end{array}\\right)=\\frac{9!}{7!\\left(9 - 7\\right)!}=\\frac{9\\cdot 8\\cdot 7!}{7!2!}=36[\/latex]<\/li>\n<\/ol><\/div>\n
      \n

      Analysis of the Solution<\/h3>\nNotice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.\n
      [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=\\left(\\begin{array}{c}n\\\\ n-r\\end{array}\\right)[\/latex]<\/div>\n<\/div>\n
      \n

      Try It 1<\/h3>\nFind each binomial coefficient.\n
      a. [latex]\\left(\\begin{array}{c}7\\\\ 3\\end{array}\\right)[\/latex]<\/div>\n
      b. [latex]\\left(\\begin{array}{c}11\\\\ 4\\end{array}\\right)[\/latex]<\/div>\nSolution<\/a>\n\n<\/div>","rendered":"

      In Counting Principles<\/a>, we studied combinations<\/strong>. In the shortcut to finding [latex]{\\left(x+y\\right)}^{n}[\/latex], we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)[\/latex] instead of [latex]C\\left(n,r\\right)[\/latex], but it can be calculated in the same way. So\n<\/p>\n

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

      The combination [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)[\/latex] is called a binomial coefficient<\/strong>. An example of a binomial coefficient is [latex]\\left(\\begin{array}{c}5\\\\ 2\\end{array}\\right)=C\\left(5,2\\right)=10[\/latex].<\/p>\n

      \n

      A General Note: Binomial Coefficients<\/h3>\n

      If [latex]n[\/latex] and [latex]r[\/latex] are integers greater than or equal to 0 with [latex]n\\ge r[\/latex], then the binomial coefficient<\/strong> is<\/p>\n

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

      Q & A<\/h3>\n

      Is a binomial coefficient always a whole number?<\/h3>\n

      Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number. <\/em><\/p>\n<\/div>\n

      \n

      Example 1: Finding Binomial Coefficients<\/h3>\n

      Find each binomial coefficient.<\/p>\n

        \n
      1. [latex]\\left(\\begin{array}{c}5\\\\ 3\\end{array}\\right)[\/latex]<\/li>\n
      2. [latex]\\left(\\begin{array}{c}9\\\\ 2\\end{array}\\right)[\/latex]<\/li>\n
      3. [latex]\\left(\\begin{array}{c}9\\\\ 7\\end{array}\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n
        \n

        Solution<\/h3>\n

        Use the formula to calculate each binomial coefficient. You can also use the [latex]{n}_{}{C}_{r}[\/latex] function on your calculator.<\/p>\n

        [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\n
          \n
        1. [latex]\\left(\\begin{array}{c}5\\\\ 3\\end{array}\\right)=\\frac{5!}{3!\\left(5 - 3\\right)!}=\\frac{5\\cdot 4\\cdot 3!}{3!2!}=10[\/latex]<\/li>\n
        2. [latex]\\left(\\begin{array}{c}9\\\\ 2\\end{array}\\right)=\\frac{9!}{2!\\left(9 - 2\\right)!}=\\frac{9\\cdot 8\\cdot 7!}{2!7!}=36[\/latex]<\/li>\n
        3. [latex]\\left(\\begin{array}{c}9\\\\ 7\\end{array}\\right)=\\frac{9!}{7!\\left(9 - 7\\right)!}=\\frac{9\\cdot 8\\cdot 7!}{7!2!}=36[\/latex]<\/li>\n<\/ol>\n<\/div>\n
          \n

          Analysis of the Solution<\/h3>\n

          Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.<\/p>\n

          [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=\\left(\\begin{array}{c}n\\\\ n-r\\end{array}\\right)[\/latex]<\/div>\n<\/div>\n
          \n

          Try It 1<\/h3>\n

          Find each binomial coefficient.<\/p>\n

          a. [latex]\\left(\\begin{array}{c}7\\\\ 3\\end{array}\\right)[\/latex]<\/div>\n
          b. [latex]\\left(\\begin{array}{c}11\\\\ 4\\end{array}\\right)[\/latex]<\/div>\n

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

          \n\t\t\t

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