{"id":2102,"date":"2015-11-12T18:30:42","date_gmt":"2015-11-12T18:30:42","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=2102"},"modified":"2017-04-03T20:02:45","modified_gmt":"2017-04-03T20:02:45","slug":"solutions-11","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/solutions-11\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"

Solutions to Try Its<\/h2>\r\n1. 7\r\n\r\n2.\u00a0There are 60 possible breakfast specials.\r\n\r\n3.\u00a0120\r\n\r\n4.\u00a060\r\n\r\n5.\u00a012\r\n\r\n6.\u00a0[latex]P\\left(7,7\\right)=5,040[\/latex]\r\n\r\n7.\u00a0[latex]P\\left(7,5\\right)=2,520[\/latex]\r\n\r\n8.\u00a0[latex]C\\left(10,3\\right)=120[\/latex]\r\n\r\n9.\u00a064 sundaes\r\n\r\n10.\u00a0840\r\n

Solutions of Odd-Numbered Exercises<\/h2>\r\n1.\u00a0There are [latex]m+n[\/latex] ways for either event [latex]A[\/latex] or event [latex]B[\/latex] to occur.\r\n\r\n3.\u00a0The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word \"or\" usually implies an addition problem. The word \"and\" usually implies a multiplication problem.\r\n\r\n5.\u00a0A combination; [latex]C\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!r!}[\/latex]\r\n\r\n7.\u00a0[latex]4+2=6[\/latex]\r\n\r\n9.\u00a0[latex]5+4+7=16[\/latex]\r\n\r\n11.\u00a0[latex]2\\times 6=12[\/latex]\r\n\r\n13.\u00a0[latex]{10}^{3}=1000[\/latex]\r\n\r\n15.\u00a0[latex]P\\left(5,2\\right)=20[\/latex]\r\n\r\n17.\u00a0[latex]P\\left(3,3\\right)=6[\/latex]\r\n\r\n19.\u00a0[latex]P\\left(11,5\\right)=55,440[\/latex]\r\n\r\n21.\u00a0[latex]C\\left(12,4\\right)=495[\/latex]\r\n\r\n23.\u00a0[latex]C\\left(7,6\\right)=7[\/latex]\r\n\r\n25.\u00a0[latex]{2}^{10}=1024[\/latex]\r\n\r\n27.\u00a0[latex]{2}^{12}=4096[\/latex]\r\n\r\n29.\u00a0[latex]{2}^{9}=512[\/latex]\r\n\r\n31.\u00a0[latex]\\frac{8!}{3!}=6720[\/latex]\r\n\r\n33.\u00a0[latex]\\frac{12!}{3!2!3!4!}[\/latex]\r\n\r\n35.\u00a09\r\n\r\n37.\u00a0Yes, for the trivial cases [latex]r=0[\/latex] and [latex]r=1[\/latex]. If [latex]r=0[\/latex], then [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=1\\text{.}\\hspace{0.17em}[\/latex] If [latex]r=1[\/latex], then [latex]r=1[\/latex], [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=n[\/latex].\r\n\r\n39.\u00a0[latex]\\frac{6!}{2!}\\times 4!=8640[\/latex]\r\n\r\n41.\u00a0[latex]6 - 3+8 - 3=8[\/latex]\r\n\r\n43.\u00a0[latex]4\\times 2\\times 5=40[\/latex]\r\n\r\n45.\u00a0[latex]4\\times 12\\times 3=144[\/latex]\r\n\r\n47.\u00a0[latex]P\\left(15,9\\right)=1,816,214,400[\/latex]\r\n\r\n49.\u00a0[latex]C\\left(10,3\\right)\\times C\\left(6,5\\right)\\times C\\left(5,2\\right)=7,200[\/latex]\r\n\r\n51.\u00a0[latex]{2}^{11}=2048[\/latex]\r\n\r\n53.\u00a0[latex]\\frac{20!}{6!6!8!}=116,396,280[\/latex]","rendered":"

Solutions to Try Its<\/h2>\n

1. 7<\/p>\n

2.\u00a0There are 60 possible breakfast specials.<\/p>\n

3.\u00a0120<\/p>\n

4.\u00a060<\/p>\n

5.\u00a012<\/p>\n

6.\u00a0[latex]P\\left(7,7\\right)=5,040[\/latex]<\/p>\n

7.\u00a0[latex]P\\left(7,5\\right)=2,520[\/latex]<\/p>\n

8.\u00a0[latex]C\\left(10,3\\right)=120[\/latex]<\/p>\n

9.\u00a064 sundaes<\/p>\n

10.\u00a0840<\/p>\n

Solutions of Odd-Numbered Exercises<\/h2>\n

1.\u00a0There are [latex]m+n[\/latex] ways for either event [latex]A[\/latex] or event [latex]B[\/latex] to occur.<\/p>\n

3.\u00a0The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word “or” usually implies an addition problem. The word “and” usually implies a multiplication problem.<\/p>\n

5.\u00a0A combination; [latex]C\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!r!}[\/latex]<\/p>\n

7.\u00a0[latex]4+2=6[\/latex]<\/p>\n

9.\u00a0[latex]5+4+7=16[\/latex]<\/p>\n

11.\u00a0[latex]2\\times 6=12[\/latex]<\/p>\n

13.\u00a0[latex]{10}^{3}=1000[\/latex]<\/p>\n

15.\u00a0[latex]P\\left(5,2\\right)=20[\/latex]<\/p>\n

17.\u00a0[latex]P\\left(3,3\\right)=6[\/latex]<\/p>\n

19.\u00a0[latex]P\\left(11,5\\right)=55,440[\/latex]<\/p>\n

21.\u00a0[latex]C\\left(12,4\\right)=495[\/latex]<\/p>\n

23.\u00a0[latex]C\\left(7,6\\right)=7[\/latex]<\/p>\n

25.\u00a0[latex]{2}^{10}=1024[\/latex]<\/p>\n

27.\u00a0[latex]{2}^{12}=4096[\/latex]<\/p>\n

29.\u00a0[latex]{2}^{9}=512[\/latex]<\/p>\n

31.\u00a0[latex]\\frac{8!}{3!}=6720[\/latex]<\/p>\n

33.\u00a0[latex]\\frac{12!}{3!2!3!4!}[\/latex]<\/p>\n

35.\u00a09<\/p>\n

37.\u00a0Yes, for the trivial cases [latex]r=0[\/latex] and [latex]r=1[\/latex]. If [latex]r=0[\/latex], then [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=1\\text{.}\\hspace{0.17em}[\/latex] If [latex]r=1[\/latex], then [latex]r=1[\/latex], [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=n[\/latex].<\/p>\n

39.\u00a0[latex]\\frac{6!}{2!}\\times 4!=8640[\/latex]<\/p>\n

41.\u00a0[latex]6 - 3+8 - 3=8[\/latex]<\/p>\n

43.\u00a0[latex]4\\times 2\\times 5=40[\/latex]<\/p>\n

45.\u00a0[latex]4\\times 12\\times 3=144[\/latex]<\/p>\n

47.\u00a0[latex]P\\left(15,9\\right)=1,816,214,400[\/latex]<\/p>\n

49.\u00a0[latex]C\\left(10,3\\right)\\times C\\left(6,5\\right)\\times C\\left(5,2\\right)=7,200[\/latex]<\/p>\n

51.\u00a0[latex]{2}^{11}=2048[\/latex]<\/p>\n

53.\u00a0[latex]\\frac{20!}{6!6!8!}=116,396,280[\/latex]<\/p>\n\n\t\t\t

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