Assignment 3

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20:45:22 `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 --> Order: 12 * 11 * 10 * 9 No order: 12 * 11 *10 * 9 / (4 * 3 * 2 * 1)

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20:45:23 When choosing 4 objects out of 12, there are 12 choices for the first, 11 choices for the second, 10 choices for the third and 9 choices for the fourth object. If the order matters there are therefore 12 * 11 * 10 * 9 possible outcomes. If the order doesn't matter, then we have to ask in how many different orders any given collection of 4 objects could be chosen. Given any 4 objects, there are 4 choices for the first, 3 choices for the second, 2 choices for the third and 1 choice for the fourth. There are thus 4 * 3 * 2 * 1 orders in which a given set of 4 objects could be chosen. We therefore have 12 * 11 * 10 * 9 / ( 4 * 3 * 2 * 1) possible outcomes when order doesn't matter.

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20:45:58 `q002. If order does not matter, then how many ways are there to choose 5 members of a team from 23 potential players?

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RESPONSE --> 23 * 22 * 21* 20 * 19 / (5 * 4 * 3 * 2* 1)

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20:46:02 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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20:46:42 `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

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20:46:58 It should be clear that there are 5 * 4 * 3 * 2 * 1 ways, since there are 5 choices for the first book, 4 for the second, etc.. If we multiply these numbers out we get 5 * 4 * 3 * 2 * 1 = 120. It might be a little bit surprising that there should be 120 ways to order only 5 objects.

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20:48:33 `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 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

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20:49:05 6 ! = 6 * 5 * 4 * 3 * 2 * 1 = 720. 7 ! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040. 10 ! = 3,628,800. These numbers grow at an astonishing rate. The last result here shows is that there are over 3 million ways to arrange 10 people in a line. The rapid growth of these results like in part explain the use of the ! symbol to designate factorials.

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21:10:20 `q005. What do we get if we simplify the expression (10 ! / 6 !) ?

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RESPONSE --> 10 * 9 * 8 * 7 = 5040

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21:10:54 10 ! / 6 ! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 6 * 5 * 4 * 3 * 2 * 1). We can simplify this by rewriting it as 10 * 9 * 8 * 7 * (6 * 5 * 4 * 3 * 2 * 1) / ( 6 * 5 * 4 * 3 * 2 * 1) = 10 * 9 * 8 * 7. We see that the 6 * 5 * 4 * 3 * 2 * 1 in the numerator matches the same expression in the denominator, so when divided these expressions give us 1 and we end up with just 10 * 9 * 8 * 7 * 1 = 10 * 9 * 8 * 7. Note that this is just the number of ways in which 4 objects can be chosen, in order, from a collection of 10 objects.

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21:11:24 `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 --> 23! / 5!

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21:16:20 If we divide 23 ! by 18 ! , the numbers from 18 down to 1 will occur in both the numerator and denominator and when we divide we will be left with just the numbers from 23 down to 19. Thus 23 * 22 * 21 * 20 * 19 = 23 ! / 18 !.

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RESPONSE --> I understand

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21:16:53 `q007. How could we express the number of ways to rank 20 individuals, in order, from among 100 candidates?

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RESPONSE --> P(100,20)

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21:18:08 There are 100 choices for the first candidate, 99 for the second, 98 for the third, etc.. For the 20th candidate there are 81 choices. You should convince yourself of this if you didn't see it originally. Our product is therefore 100 * 99 * 98 * ... * 81, which can be expressed as 100 ! / 80 !. We see that the denominator must be (100 - 20)! . For this example 100 - 20 represents the difference between the number of individuals available and the number selected.

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RESPONSE -->

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21:18:30 `q008. How could we express the number of ways to rank r individuals from a collection of n candidates?

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RESPONSE --> C(n,r)

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21:19:38 By analogy with the preceding example, we should divide n ! by ( n - r ) !. The number is therefore n ! / ( n - r ) !.

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RESPONSE --> I tried to oversimply

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21:22:09 `q009. The expression n ! / ( n - r ) ! denotes the number of ways in which r objects can be chosen, in order, from among n objects. When we choose objects in order we say that we are 'permuting' the objects. The expression n ! / ( n - r ) ! is therefore said to the the number of permutations of r objects chosen from n possible objects. We use the notation P ( n , r ) to denote this number. Thus P(n, r) = n ! / ( n - r ) ! . Find P ( 8, 3) and explain what this number means.

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RESPONSE --> P(8-3) = 8!/ (8-3)! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / 5* 4* 3 * 2 * 1 = 8 * 7* 6 * 1 = 336 possible ways to arrange 3 objects from 8

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21:22:21 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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21:24:05 `q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates?

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RESPONSE --> C(8,3) = P(8,3) / 3! = 8!/ (8-3)! (3!) = 8*7*6*5*4*3*2*1 / 5*4*3*2*1 (3*2*1) = 4*7*2*1 = 56

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21:24:12 There are 8 * 7 * 6 ways to choose 3 objects from 8, in order, and 3! ways to order any unordered collection of 3 objects, so there are 8 * 7 * 6 / ( 3 * 2 * 1 ) possible unordered collections. This number is easily enough calculated. Since 3 goes into 6 twice and 2 goes into 8 four times, we see that 8 * 7 * 6 / ( 3 * 2 * 1) = 4 * 7 * 2 = 56. There are 56 different unordered collections of 3 objects chosen from 8.

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21:24:51 `q010. How could the result of the preceding problem be expressed purely in terms of factorials?

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RESPONSE --> 8! / 6! (8*7*6)/ (3*2*1)

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21:25:02 The product 8 * 7 * 6 is just 8 ! / 5 !, and the expression 8 * 7 * 6 / ( 3 * 2 * 1) can therefore be expressed as 8 ! / ( 5 ! * 3 !).

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21:27:28 `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 --> (16-5)!

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21:28:16 There are 16 ! / ( 16 - 5) ! Possible ordered sets of 5 objects chosen from the 16. There are 5 ! ways to order any unordered collection of 5 objects. There are thus 16 ! / [ ( 16 - 5 ) ! * 5 ! ] possible unordered collections of 5 objects from the 16.

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RESPONSE --> ok

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21:29:41 `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 --> n ! / [ (n - r ) ! * r ! ]

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21:30:11 There are P(n, r) = n ! / ( n - r ) ! possible ordered collections of r objects. There are r ! ways to order any unordered collection of r objects. There are thus P ( n, r ) / r! = n ! / [ r ! * ( n - r) ! ] possible unordered collections of r objects chosen from n objects.

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RESPONSE --> I simply applied the logic from the previous question

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21:31:19 `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 --> ok

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Good. Let me know if you have questions.