Example:
1. Problem: 4
E (2k + 1)
k = 1
Solution: This is a sum of (2k + 1) from 1 to 4.
Plug all numbers from 1 to 4 into the general term
((2k + 1) in this case) and then add the terms together.
(2(1) + 1) = 3
(2(2) + 1) = 5
(2(3) + 1) = 7
(2(4) + 1) = 9
3 + 5 + 7 + 9 = 24
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pepperoni, sausage, onions, mushrooms
sausage, onions, mushrooms, pepperoni
onions, mushrooms, pepperoni, sausage
mushrooms, pepperoni, sausage, onions
pepperoni, sausage, mushrooms, onions
There are some more, but we won't list them. To find the number of different arrangements of the set we select a first choice; there are 4 possible choices. Now we take a second choice; there are 3 choices. Now pick a third choice; there are 2 choices. Finally, there is 1 choice for the last selection. Thus, there are 4 * 3 * 2 * 1 or 24 different ordered arrangements of the toppings. This product can also be written as 4! (read: 4-factorial).
The toal number of permutations of a set of n objects is given by n!.
Example:
1. Problem: 5!
Solution: 5 * 4 * 3 * 2 * 1
120
When you have a set of objects and only want to arrange part of
them, you have a permutation of n objects r at a
time. For example, if you have 6 toppings for a pizza, and
a customer calls and tells you to put any 3 toppings on the
pizza, you might want to know how many different pizzas you can
make. You can select the first topping in 6 ways, the second
in 5, and the third in 4. As we learned above, this can be
written as 6 * 5 * 4. There is a theorem that tells
us about a formula for the situation above. It says the
number of permutations of a set of n objects taken r
at a time is given by the following formula: nPr = (n!)/(n - r)!.
Example:
1. Problem: If a school has lockers with 50 numbers on each combination
lock, how many possible combinations using three numbers are
there.
Solution: Recognize that n, or the number of objects is 50
and that r, or the number of objects taken at one time is
3.
Plug those numbers in the permutation formula.
50!
50P3 = ---------
(50 - 3)!
Use a calculator to find the final answer.
117600
Things are immensely simplified when you can repeat the
objects. For example, if you are making license
plates with only 4 letters on them, and you can repeat
the letters, you can take the first letter from 26
options, the same for the second, third, and fourth. Therefore,
there are 26^4 or
456976 available license plates using 4 letters
if you can repeat letters. There is a special
theorem that tells us the number of arrangements of
n objects taken r at a time, with repetition
is given by n^r.
Example:
1. Problem: How many 4 digit license plates can you make using the numbers
from 0 to 9 while allowing repetitions.
Solution: Realize there are 10 objects taken 4 at a time.
Plug that information into the formula for repeated use.
10^4
10000
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The number of combinations of a set of n objects taken r at a time is given by nCr = (n!)/(r!(n - r)!).
Example:
1. Problem: For a study, 4 people are chosen at random from a group of 10
people. How many ways can this be done?
Solution: Since you're going to have the same group of people no matter
the order you choose the people in, you set up the problem as a
combination.
10!
10C4 = -----------
4!(10 - 4)!
Use a calculator to find the answer.
There are 210 different groups of people you can
choose.
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Example:
1. Problem: What is the probability of rolling a 3 on a die (plural,
dice).
Solution: On a fair die (not the kind you play with in Vegas, where
everything is rigged), there are six equally likely outcomes
when you roll. Also, there is only one way to get a 3.
By the definition of probability, P(3) = (1/6).
Probability will always be a fraction, 0, or 1. If
an event cannot happen, the
probability is 0. If an event is certain to
happen, the probability is 1.
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Take the quiz on probability. The quiz is very useful for either review or to see if you've really got the topic down.