Numbers and Operations Study Guide for the Math Basics
Page 4
Factorials, Permutations and Combinations
These types of terms show up most often in tests or sections dealing with Probability and Statistics.
Factorials
If you see the expression 4!, instead of thinking 4 is very excited, realize that it means 4 factorial. This means to multiply 4 by every number lower than it, ending at 1.
\[4!=4 \cdot 3 \cdot 2 \cdot 1=24\] \[6!=6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=720\]Note: \(0! = 1\).
Permutations
A permutation (\(P\,^n_k\) or \(P(n,k)\)) is the number of ways you can re-order certain values (n) in a list of a given size (k). Imagine you are to find out how many ways A, B, and C could be listed in groups of 3 (order matters). Here are the ways:
ABC
ACB
BAC
BCA
CAB
CBA
There are 6 total ways to arrange a set of 3 elements into lists of length 3. We could answer this question without actually listing it using the formula:
\(P\,^n_k\) or \(P(n,k) = \frac{n!}{(n-k)!}\)
If the number of elements is the same as the number in each list, the formula simplifies to:
\(P\,^n_n = n!\). So, \(P\,^3_3 = 3! = 3 \cdot 2 \cdot 1 = 6\)
How many ways are there to list 5 different elements into groups of 3?
\(P(5, 3)\) or \(P\,^5_3 =\)
\[\frac{5!}{(5-3)!} = \frac{5!}{2!}\] \[\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1} = \frac{120}{2} = 60\]Combinations
Now imagine you want to find out how many ways to make groups of 3 out of 5 different people in your class. In this case, the order of the list doesn’t matter. Let’s call the people A, B, C, D, and E. Now ABC is a different list than ACB, but as a group they are the same. Combinations deal with this type of question (where order doesn’t matter).
\[C(n,k) = \frac{n!}{(n-k)! \cdot k!}\]In this case:
\[C(5,3) = \frac{5!}{(5-3)! \cdot 3!}\] \[\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} = \frac{120}{12} = 10\]So there are 10 ways to make groups of 3 out of 5 people.
Factors and Multiples
It’s very important to know the multiplication facts from 1-9 for many reasons: it helps speed up multiplication, it makes division possible, and it helps find factors and multiples.
Factors
Factors are the building blocks of numbers. You’ve already been introduced to this term when you learned multiplication: factor \(\cdot\) factor = product. For example, \(3 \cdot 4 = 12\), so \(3\) and \(4\) are factors of \(12\).
Finding Factors
To factor (a verb) a number means to write it as a product of two or more numbers: its factors (noun).
To find factors of a number, find all the numbers which divide evenly into it. Let’s find the factors of 12. First, think about (or list) all numbers between (and including) 1 and 12.
\[1, \;\;2, \;\;3,\; \;4,\;\; 5,\; \;6,\; \;7,\; \;8,\;\; 9, \;\;10,\;\;11,\;\;12\]Work your way through the list, determining if each number is a factor of (divides evenly into) 12. 1 is the first factor because \(1 \cdot 12 = 12\). This also means 12 is a factor. Make a note of this on your list (underline or circle).
\[\underline1, \;\;2, \;\;3,\; \;4,\;\; 5,\; \;6,\; \;7,\; \;8,\;\; 9, \;\;10,\;\;11,\;\; \underline{12}\]Now, keep going:
\(2\) is a factor \((2 \cdot 6 = 12)\), therefore so is \(6\).
\(3\) is a factor \((3 \cdot 4 = 12)\), and so is \(4\).
Your list should look like this now:
You know you’re finished because the next number to try would be 4, which is already a factor. So the factors of 12 are 1, 2, 3, 4, 6, and 12.
Prime Factorization
Every number has a unique way to represent it as a product of prime (and only prime) factors. Perhaps this is one reason why mathematicians seem to have an unhealthy obsession with prime numbers. To find the prime factorization of any number, write the number as a product of any pair of its factors. Check to see if either of the two factors is prime. If it is, circle it . Then, continue working down until all factors are prime. Here’s the prime factorization of 12.
\[12\] \[\require{enclose} \enclose{circle}{3} \cdot 4\] \[\quad\quad\;\enclose{circle}{2} \cdot \enclose{circle}{2}\]So the prime factorization of \(12\) is \(2 \cdot 2 \cdot 3\), or the product of all circled numbers. Let’s try it using different factors to begin with and notice that the prime factorization should be the same.
\[12\] \[\enclose{circle}{2} \cdot 6\] \[\quad\quad\;\enclose{circle}{2} \cdot \enclose{circle}{3}\]Using Factors
One great way to use factors is to break up a multiplication or division problem and make it easier for mental calculations. For a multiplication example, find:
\[45 \cdot 12\]\(12\) can be factored as \(2 \cdot 6\), so instead of finding \(45 \cdot 12\), you can find:
\[45 \cdot 2 \cdot 6 = 90 \cdot 6 = 540\]For a division example, find:
\[120 \div 15\]Since \(15\) can be written as \(3 \cdot 5\), then:
\[120 \div 15 = 120 \div (3 \cdot 5)\]When dividing a number by a product of 2 numbers, just divide it by one and then the other. So:
\[120 \div (3 \cdot 5) = (120 \div 3) \div 5 = 40 \div 5 = 8\]Common Factors and GCF
The numbers 12 and 15 have some factors in common: 1 and 3. So 1 and 3 are the common factors of 12 and 15.
More importantly, you’ll probably be asked to find the greatest common factor (GCF) between two numbers. This is helpful in many cases: simplifying fractions, factoring polynomials, and some word problems. One way is to just find all the factors of the two numbers, look for the common factors, and then choose the greatest one. Find the GCF of 24 and 66:
\(24: 1, 2, 3, 4, 6, 8, 12, 24\)
\(66: 1, 2, 3, 6, 11, 22, 33, 66\)
Common factors: \(1, 2, 3, 6\)
GCF \((24, 66) = 6\)
This process can seem time-consuming, especially for big numbers. Another method is to find the prime factorization of each number. The GCF will be the product of all the common factors.
\[\begin{array}{|c|c|} \hline 24 & 66 \\ {\enclose{circle}{2} \cdot 12} & {\enclose{circle}{11} \cdot 6} \\ {\quad\;\enclose{circle}{2} \cdot 6} & {\quad\quad\quad\quad\;\enclose{circle}{2} \cdot \enclose{circle}{3}\quad\quad} \\ \quad\quad\quad\enclose{circle}{3} \cdot \enclose{circle}{3}\\ \hline \end{array}\]\(24 = 2 \cdot 2 \cdot 2 \cdot 3\)
\(66 = 2 \cdot 3 \cdot 11\)
Common Prime Factors: \(2, 3\)
GCF \((24,66) = 2 \cdot 3 = 6\)
Sometimes, two numbers have multiple copies of the same prime factor in common. Here’s an example showing how that works.
\(24 = 2 \cdot 2 \cdot 2 \cdot 3\)
\(36 = 2 \cdot 2 \cdot 3 \cdot 3\)
Common Prime Factors: \(2, 2, 3\)
GCF \((24,36) = 2 \cdot 2 \cdot 3 = 12\)
Multiples
You already know what multiples are if you know your multiplication tables. 9, 18, 27, 36, etc. are all multiples of 9, for example.
Finding Multiples
To find multiples of a number, just multiply that number by 1, 2, 3, 4, and so on. Multiples of 11 are 11, 22, 33, 44, 55, …
Using Multiples
The most common use of multiples is division. Instead of looking at a division problem as repeated subtraction, speed up the process by seeing if the dividend is a multiple of the divisor. For example: \(36 \div 12 = 3\) because \(36\) is a multiple (\(3\) times) of \(12\).
Common Multiples and LCM
Let’s imagine the following scenario. You’re at the store buying hot dogs and buns for a big barbecue. Hot dogs come in packs of 10 and buns come in packs of 8 (what sadistic human being decided to do this?!). You want to buy enough hot dogs and buns so none are wasted. Perhaps you think: I’ll buy 80 hot dogs and buns (8 packs of hot dogs and 10 packs of buns). This is a possible solution because 80 is a common multiple of 8 and 10.
But maybe that’s too much. Is this the least amount you can buy without wastage? To find this amount, you’ll need to find the least common multiple (LCM). In this case, the LCM is 40 (4 packs of hot dogs and 5 packs of buns). To find the LCM, list the multiples of each number, find the common ones, and choose the smallest.
\(10: 10, 20, 30, 40, 50, 60, 70, 80, 90, …\)
\(8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, …\)
Common multiples: \(40, 80, (120), …\)
LCM \((10, 8) = 40\)
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