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Even if you're not quite sure what a "combination" is in the world of math, you've dealt with them before. We create combinations every time we choose our veggies in a sandwich or a burger. Every time we get up in the morning, we choose a different combination of clothing before heading to school. After learning more about the mathematical aspects of combinations, we can make calculations that can be very useful in life.
When you enter a certain number into a combination lock, you are creating a combination. This is a "set" of numbers that can be arranged in any way and in any order. The last point is especially important, as this sets combinations apart from permutations. While permutations require values to be arranged in a specific order, combinations do not. The easiest way to understand this concept is by looking at two concepts: Stir fry and cooking pasta.
When we cook up a stir-fry, the order doesn't really matter. For the most part, we can get away with tossing our ingredients into a pan (or a wok) at the same time and stirring everything until our meal is ready. Because the order does not matter, this is a combination.
On the other hand, the same concept does not apply to cooking pasta. We know that we need to add the water first, bring it to a boil, and then add the pasta. The order matters, so this is a permutation and not a combination.
Note that the word "permutation" isn't very common in everyday conversation. This is why we call certain padlocks "combination locks," despite the fact that the numbers must be arranged in a specific manner. Technically, the correct term should be "permutation lock."
Under the umbrella term of "combination," there are two more specific categories to consider:
Because order does not matter when we deal with combinations, we might argue that they are simpler versions of permutations. While this is true to some extent, the calculations we need to make are actually a little more complex.
You may recall that the formula for permutations is "n choose r" or $\frac{n!}{\left(n-r\right)!}$ .
In this formula, n is the number of choices while r is the number of choices we can make. We can also write this as ${}_{n}P_{r}$ . For example, if we have 16 choices and we can choose three values per choice, the resulting expression is ${}_{16}P_{3}$ . We can solve this problem with the following operation:
$16\times 15\times 14=3360$ possibilities
Note that we used 16, 15, and 14 because the operation is factorial, $r$ is 3, so we used 16 and the next 2 integers.
We can apply this same pattern to other permutation calculations.
But we're talking about combinations -- not permutations. So how do we adjust this formula so that it helps us solve combinations?
We change our formula from:
$\frac{n!}{\left(n-r\right)!}$
To:
$\frac{n!}{r!\left(n-r\right)!}$
We call this the binomial coefficient.
In other words, we are doing our normal permutation calculation before dividing out the number of different orders for each group. This leaves us with a greater number of possible combinations compared to our permutations.
For example, if we have 16 balls with different numbers and we can only choose three balls, the formula for our combination calculation would be:
$\frac{16!}{3!\left(16-3\right)!}$
This gives us a value of 560.
But what about combinations that can repeat? This is a much more complex problem. The easiest method involves summarizing the different decisions into binary. We can imagine all of the choices displayed in a straight line. Each possible result is a combination of either "choose" or "pass." We can imagine using a specific code to program a robot to "choose" or "pass" each box in different orders. Because repetition is allowed, the robot could choose the same box multiple times.
The resulting formula would look something like this: $C(n+r-1,r)=\frac{(n+r-1)!}{\left[r!(n-1)!\right]}$ , where n is the number of types of items and r is the number of items being chosen.
For now, let's focus on combinations that cannot repeat.
Let's say we've been tasked with testing four smartphones out of a batch of 100 to see whether they're functional. Assume that each smartphone is different.
We can plug this into our formula:
${}_{n}C_{m}$
${}_{100}C_{4}$
$=\frac{100!}{4!(100-4)!}$
$=\frac{100!}{4!(96!)}$
$=3921225$
Therefore, there are 3,921,225 different combinations in this case.
Common Core: High School - Statistics and Probability Flashcards
Probability Theory Practice Tests
Common Core: High School - Statistics and Probability Diagnostic Tests
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