SECTION 13.2 Permutations and Combinations 921 Using Formula (2) Use formula (2) to find the value of each combination. (a) ( ) C 3, 1 (b) ( ) C 6, 3 (c) ( ) C n n , (d) ( ) C n, 0 (e) ( ) C 52, 5 Solution EXAMPLE 7 (a) ( ) ( ) = − = = ⋅ ⋅ ⋅ ⋅ = C 3, 1 3! 3 1 ! 1! 3! 2! 1! 3 2 1 2 1 1 3 (b) C 6, 3 6! 6 3 ! 3! 6 5 4 3! 3! 3! 6 5 4 6 20 ( ) ( ) = − = ⋅ ⋅ ⋅ = ⋅ ⋅ = (c) ( ) ( ) = − = = = C n n n n n n n n , ! ! ! ! 0! ! 1 1 1 (d) ( ) ( ) = − = = = C n n n n n , 0 ! 0 ! 0! ! !0! 1 1 1 (e) Figures 4(a) and 4(b) show the solution using a TI-84 Plus CE graphing calculator and Desmos. ( ) = C 52, 5 2,598,960 Figure 4(a) ( ) C 52, 5 Figure 4(b) Now Work PROBLEM 15 Forming Committees How many different committees of 3 people can be formed from a group of 7 people? Solution EXAMPLE 8 The 7 people are distinct. More important, though, is the observation that the order of being selected for a committee is not significant.The problem asks for the number of combinations of 7 objects taken 3 at a time. ( ) = = ⋅ ⋅ ⋅ = ⋅ ⋅ = C 7, 3 7! 4! 3! 7 6 5 4! 4! 3! 7 6 5 6 35 Thirty-five different committees can be formed. Forming Committees In how many ways can a committee consisting of 2 faculty members and 3 students be formed if 6 faculty members and 10 students are eligible to serve on the committee? Solution EXAMPLE 9 The problem can be separated into two parts: the number of ways in which the faculty members can be chosen, ( ) C 6, 2 , and the number of ways in which the student members can be chosen, ( ) C 10, 3 . By the Multiplication Principle, the committee can be formed in ( ) ( ) ⋅ = ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ = C C 6, 2 10, 3 6! 4! 2! 10! 7! 3! 6 5 4! 4! 2! 10 9 8 7! 7! 3! 30 2 720 6 1800 ways Now Work PROBLEM 49
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