# 9.5 Counting principles  (Page 5/12)

 Page 5 / 12
$\text{\hspace{0.17em}}C\left(5,0\right)+C\left(5,1\right)+C\left(5,2\right)+C\left(5,3\right)+C\left(5,4\right)+C\left(5,5\right)=32\text{\hspace{0.17em}}$

There are 32 possible pizzas. This result is equal to $\text{\hspace{0.17em}}{2}^{5}.\text{\hspace{0.17em}}$

We are presented with a sequence of choices. For each of the $n$ objects we have two choices: include it in the subset or not. So for the whole subset we have made $n\text{\hspace{0.17em}}$ choices, each with two options. So there are a total of $2·2·2·\dots ·2$ possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time.

## Formula for the number of subsets of a set

A set containing n distinct objects has ${2}^{n}$ subsets.

## Finding the number of subsets of a set

A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?

We are looking for the number of subsets of a set with 4 objects. Substitute $n=4$ into the formula.

There are 16 possible ways to order a potato.

A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?

64 sundaes

## Finding the number of permutations of n Non-distinct objects

We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be $12!$ ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the $12!$ permutations we counted are duplicates. The general formula for this situation is as follows.

$\text{\hspace{0.17em}}\frac{n!}{{r}_{1}!{r}_{2}!\dots {r}_{k}!}\text{\hspace{0.17em}}$

In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are $4!$ ways to order the stars and $3!$ ways to order the moon.

$\text{\hspace{0.17em}}\frac{12!}{4!3!}=3\text{,}326\text{,}400\text{\hspace{0.17em}}$

There are 3,326,400 ways to order the sheet of stickers.

## Formula for finding the number of permutations of n Non-distinct objects

If there are $n$ elements in a set and ${r}_{1}\text{\hspace{0.17em}}$ are alike, $\text{\hspace{0.17em}}{r}_{2}\text{\hspace{0.17em}}$ are alike, ${r}_{3}\text{\hspace{0.17em}}$ are alike, and so on through ${r}_{k},\text{\hspace{0.17em}}$ the number of permutations can be found by

$\text{\hspace{0.17em}}\frac{n!}{{r}_{1}!{r}_{2}!\dots {r}_{k}!}\text{\hspace{0.17em}}$

## Finding the number of permutations of n Non-distinct objects

Find the number of rearrangements of the letters in the word DISTINCT.

There are 8 letters. Both I and T are repeated 2 times. Substitute and into the formula.

There are 10,080 arrangements.

Find the number of rearrangements of the letters in the word CARRIER.

840

Access these online resources for additional instruction and practice with combinations and permutations.

## Key equations

 number of permutations of $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ distinct objects taken $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ at a time $P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$ number of combinations of $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ distinct objects taken $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ at a time $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ number of permutations of $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ non-distinct objects $\frac{n!}{{r}_{1}!{r}_{2}!\dots {r}_{k}!}$

## Key concepts

• If one event can occur in $m$ ways and a second event with no common outcomes can occur in $n$ ways, then the first or second event can occur in $m+n$ ways. See [link] .
• If one event can occur in $m$ ways and a second event can occur in $n$ ways after the first event has occurred, then the two events can occur in $m×n$ ways. See [link] .
• A permutation is an ordering of $n$ objects.
• If we have a set of $n$ objects and we want to choose $r$ objects from the set in order, we write $P\left(n,r\right).$
• Permutation problems can be solved using the Multiplication Principle or the formula for $P\left(n,r\right).$ See [link] and [link] .
• A selection of objects where the order does not matter is a combination.
• Given $n$ distinct objects, the number of ways to select $r$ objects from the set is $\text{C}\left(n,r\right)$ and can be found using a formula. See [link] .
• A set containing $n$ distinct objects has ${2}^{n}$ subsets. See [link] .
• For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See [link] .

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar