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course mth 152
9/15 11
Question: `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 possible 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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
12 * 11 * 10 * 9 / ( 4 * 3 * 2 * 1) possible outcomes when order doesn't matter
3
Ok
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Question:
`q002. If order does not matter, then how many ways are there to choose 5 members of a team from 23 potential players?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
23 * 22 * 21 * 20 * 19 / ( 5 * 4 * 3 * 2 * 1) possible 5-member teams.
3
Ok
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Question:
`q003. In how many ways can we line up 5 different books on a shelf?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Its probably even more surprising that if we double the number of objects to 10, there are over 3 million ways to order them
3
Ok
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Question:
`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 ! .
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
6 ! = 6 * 5 * 4 * 3 * 2 * 1 = 720. 7 ! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040. 10 ! = 3,628,800.
3
Ok
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Question:
`q005. What do we get if we simplify the expression (10 ! / 6 !) ?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
10 * 9 * 8 * 7 * 1 = 10 * 9 * 8 * 7
3
Ok
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Question:
`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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
23 * 22 * 21 * 20 * 19 = 23 ! / 18 !
3
Ok
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Question:
`q007. How could we express the number of ways to rank 20 individuals, in order, from among 100 candidates?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
100 ! / (100 20) ! = 100 ! / 80 ! = 100 * 99 * 98 *
* 81
3
Ok
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Question:
`q008. How could we express the number of ways to rank r individuals from a collection of n candidates?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
n ! / ( n - r ) !
3
Ok
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Question:
`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 be 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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6
3
Ok
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Question:
`q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
8 * 7 * 6 / ( 3 * 2 * 1) = 4 * 7 * 2 = 56
2
Ok
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Question:
`q010. How could the result of the preceding problem be expressed purely in terms of factorials?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
8 ! / ( 5 ! * 3 !)
3
Ok
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Question:
`q011. In terms of factorials, how would we express the number of possible unordered collections of 5 objects chosen from 16?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
16 ! / [ ( 16 - 5 ) ! * 5 ! ]
3
Ok
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Question:
`q012. In terms of factorials, how would we express the number of possible unordered collections of r objects chosen from n objects?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
P ( n, r ) / r! = n ! / [ r ! * ( n - r) ! ]
3
Ok
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Question: g
`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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your Solution:
C ( n , r ) = P ( n, r) / r! = n ! / [ r ! ( n - r) ! ]
3
ok
"
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