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A family of five is having portraits taken. Use the Multiplication Principle to find the following.
How many ways can the photographer line up 3 family members?
60
How many ways can the family line up for the portrait if the parents are required to stand on each end?
12
For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set of $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ objects and we want to choose $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ objects from the set in order, we write $\text{\hspace{0.17em}}P(n,r).\text{\hspace{0.17em}}$ Another way to write this is ${n}_{}{P}_{r},\text{\hspace{0.17em}}$ a notation commonly seen on computers and calculators. To calculate $\text{\hspace{0.17em}}P(n,r),\text{\hspace{0.17em}}$ we begin by finding $\text{\hspace{0.17em}}n!,\text{\hspace{0.17em}}$ the number of ways to line up all $n$ objects. We then divide by $\text{\hspace{0.17em}}\left(n-r\right)!\text{\hspace{0.17em}}$ to cancel out the $\text{\hspace{0.17em}}\left(n-r\right)\text{\hspace{0.17em}}$ items that we do not wish to line up.
Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is $6\times 5\times 4=120.$ Using factorials, we get the same result.
There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows.
Note that the formula stills works if we are choosing all $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ objects and placing them in order. In that case we would be dividing by $\text{\hspace{0.17em}}\left(n-n\right)!\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}0!,\text{\hspace{0.17em}}$ which we said earlier is equal to 1. So the number of permutations of $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ objects taken $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ at a time is $\text{\hspace{0.17em}}\frac{n!}{1}\text{\hspace{0.17em}}$ or just $\text{\hspace{0.17em}}n!\text{.}$
Given $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ distinct objects, the number of ways to select $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ objects from the set in order is
Given a word problem, evaluate the possible permutations.
A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions?
Substitute $\text{\hspace{0.17em}}n=12\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=9\text{\hspace{0.17em}}$ into the permutation formula and simplify.
There are 79,833,600 possible permutations of exam questions!
Could we have solved [link] using the Multiplication Principle?
Yes. We could have multiplied $\text{\hspace{0.17em}}15\cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\text{\hspace{0.17em}}$ to find the same answer .
A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following.
How many ways can the 7 actors line up?
$\text{\hspace{0.17em}}P(7,7)=5,040\text{\hspace{0.17em}}$
How many ways can 5 of the 7 actors be chosen to line up?
$\text{\hspace{0.17em}}P(7,5)=2,520\text{\hspace{0.17em}}$
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