course Mth 151 Sorry so late!! YO\assignment #002
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21:51:13 `questionNumber 30000 `q001. Note that there are 13 questions in this assignment. As we have seen if we choose, say, 3 objects out of 10 distinct objects the number of possible results depends on whether order matters or not. For the present example if order does matter there are 10 choices for the first selection, 9 for the second and 8 for the third, giving us 10 * 9 * 8 possibilities. However if order does not matter then whatever three objects are chosen, they could have been chosen in 3 * 2 * 1 = 6 different orders. This results in only 1/6 as many possibilities, or 10 * 9 * 8 / 6 possible outcomes. We usually write this number as 10 * 9 * 8 / (3 * 2 * 1) in order to remind us that there are 10 * 9 * 8 ordered outcomes, but 3 * 2 * 1 orders in which any three objects can be chosen. If we were to choose 4 objects out of 12, how many possibile outcomes would there be if the objects were chosen in order? How many possible outcomes would there be if the order of the objects did not matter?
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RESPONSE --> 12*11*10*9 possible outcomes 12*11*10*9 4*3*2*1 confidence assessment: 2
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21:55:21 `questionNumber 30000 If order did matter then there would be 23 * 22 * 21 * 20 * 19 ways choose the five members. However order does not matter, so we must divide this number by the 5 * 4 * 3 * 2 * 1 ways in which any given set of five individuals can be chosen. We therefore have 23 * 22 * 21 * 20 * 19 / ( 5 * 4 * 3 * 2 * 1) possible 5-member teams.
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RESPONSE --> 23*22*21*20*19/ 5*4*3*2*1 possible tem members self critique assessment: 3
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21:56:43 `questionNumber 30000 `q003. In how many ways can we line up 5 different books on a shelf?
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RESPONSE --> 5*4*3*2*1=120 ways to arange books on a shelf confidence assessment: 3
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22:01:04 `questionNumber 30000 `q004. The expression 5 * 4 * 3 * 2 * 1 is often written as 5 ! , read 'five factorial'. More generally if n stands for any number, then n ! stands for the number of ways in which n distinct objects could be lined up. Find 6 ! , 7 ! and 10 ! .
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RESPONSE --> 6! = 6*5*4*3*2*1 =720 7! = 7*6*5*4*3*2*1 = 5040 10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800 confidence assessment: 3
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22:03:15 `questionNumber 30000 `q005. What do we get if we simplify the expression (10 ! / 6 !) ?
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RESPONSE --> (10!/6!) 10*9*8*7*(6*5*4*3*2*1)/(6*5*4*3*2*1) = 10*9*8*7 confidence assessment: 2
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22:22:20 `questionNumber 30000 `q006. We saw above that there are 23 * 22 * 21 * 20 * 19 ways to choose 5 individuals, in order, from 23 potential members. How could we express this number as a quotient of two factorials?
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RESPONSE --> we can express this as a two factorial by dividing 23! by 18! on one side of the problem, and on the other side divide 23! by 19! confidence assessment: 3
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22:24:41 `questionNumber 30000 `q007. How could we express the number of ways to rank 20 individuals, in order, from among 100 candidates?
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RESPONSE --> 100*99*98*97*96*95*94*93*92*91..............................*81 = 100!/81! confidence assessment: 2
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22:26:23 `questionNumber 30000 `q008. How could we express the number of ways to rank r individuals from a collection of n candidates?
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RESPONSE --> You must divide n!/(n-r)! n!/ (n-r)! confidence assessment: 1
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22:30:27 `questionNumber 30000 P(n, r) = n! / ( n - r) !. To calculate P(8, 3) we let n = 8 and r = 3. We get P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6. This number represents the number of ways 3 objects can be chosen, in order, from 8 objects.
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RESPONSE --> n=8 r=3 8(8,3)= 8! / (8-3)! = 8!/ 5! =8*7*6*5*4*3*2*1 / (5*4*3*2*1) = 8*7*6 self critique assessment: 2
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22:32:34 `questionNumber 30000 `q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates?
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RESPONSE --> 8*7*6/ (3*2*1) = 4*7*2 = 56 There are 56different unordered collections of 3 objects from 8 confidence assessment: 3
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22:35:01 `questionNumber 30000 `q010. How could the result of the preceding problem be expressed purely in terms of factorials?
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RESPONSE --> 8*7*6/ (3*2*1) 8! / (5!*3!) confidence assessment: 1
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22:37:19 `questionNumber 30000 `q011. In terms of factorials, how would we express the number of possible unordered collections of 5 objects chosen from 16?
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RESPONSE --> 16! ? (16-5)! 16! /[(16-5)! *5!] confidence assessment: 2
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22:41:03 `questionNumber 30000 `q012. In terms of factorials, how would we express the number of possible unordered collections of r objects chosen from n objects?
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RESPONSE --> p(n,r) /r! =n!/ [r!*(n-r)!] confidence assessment: 2
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22:46:16 `questionNumber 30000 `q013. When we choose objects without regard to order, we say that we are forming combinations as opposed to permutations, which occur when order matters. The expression we obtained in the preceding problem gives us a formula for combinations: C ( n , r ) = P ( n, r) / r! = n ! / [ r ! ( n - r) ! ] gives us to number of possible combinations, or unordered collections, of r objects chosen from a set of n objects.
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RESPONSE --> c(n-r)=P(n,r)/ r!=n!/ [r!/[r!/[r!(r!(n-r)!] confidence assessment: 2
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w|Iϴp^ assignment #004 004. Dice, trees, committees, number of subsets. Liberal Arts Mathematics II 03-13-2008
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20:58:28 `questionNumber 40000 `q001. Note that there are 9 questions in this assignment. In how many ways can we get a total of 9 when rolling two fair dice?
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RESPONSE --> 4 pairs (6,3) , (5,4) , (4,5) , (3,6) confidence assessment: 3
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21:03:50 `questionNumber 40000 `q002. In how many ways can we choose a committee of three people from a set of five people?
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RESPONSE --> 10 possible 3 members within a group of 5. c(5,3)=5! /[3! (5-3)!] = 5![3!*2!]= 5*4*3*2*1/[(3*2*1)*(2*1)]= 5*4/(2*1)= 5*2= 10 confidence assessment: 2
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21:09:34 `questionNumber 40000 `q003. In how many ways can we choose a president, a secretary and a treasurer from a group of 10 people?
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RESPONSE --> p(10,3) = 10! / (10-3)! =10!/ 7!=10*9*8=720 confidence assessment: 2
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21:10:46 `questionNumber 40000 `q004. In how many ways can we arrange six people in a line?
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RESPONSE --> 6*5*4*3*2*1=6!=720 confidence assessment: 1
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21:12:17 `questionNumber 40000 `q005. In how many ways can we rearrange the letters in the word 'formed'?
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RESPONSE --> The word formed has 6 letters 6*5*4*3*2*1=720 confidence assessment: 3
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21:16:43 `questionNumber 40000 `q006. In how many ways can we rearrange the letters in the word 'activities'?
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RESPONSE --> The word activities has 10 letters. 10! 3!=6 2!=2 10! by 3!by2! 10!/ (3!*2!) confidence assessment: 3
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21:18:41 `questionNumber 40000 `q007. In how many ways can we line up four people, chosen from a group of 10, for a photograph?
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RESPONSE --> p(10,4) = 10! /(10-4)! = 10! / 6!=10*9*8*7 confidence assessment: 1
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21:20:26 `questionNumber 40000 `q008. In how many ways can we get a total greater than 3 when rolling two fair dice?
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RESPONSE --> 3 ways to get 3 or less confidence assessment: 1
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21:23:11 `questionNumber 40000 `q009. A committee consists of 5 men and 7 women. In how many ways can a subcommittee of 4 be chosen if the number of men and women must be equal?
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RESPONSE --> c(5,2) ways 2 out of 5 c(7,2) ways 2 out of 7 2 men and 2 women c(5,2)*c(7,2)= 10*21= 210 confidence assessment: 3
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