If you’re looking to understand how many unique combinations can be created from a set of objects, you’ve come to the right place. Combinations are a powerful tool utilized in a variety of ways, such as in mathematics, cryptography, and even in business. With the right understanding of how combinations work and the different kinds of combinations, you can explore a seemingly infinite number of possibilities. In this blog post, we’ll discuss the different types of combinations and the process of calculating the total number of combinations in a set of objects. We’ll also provide a few examples to help you better understand how to calculate the combinations of any set of objects. So, read on to learn more about how many combinations are possible and the process of calculating them.
Combinations made easy
Formula for finding how many combinations you have
The number of ways to create a non-repeating arrangement of items (r) from a larger set of distinct items (n) is determined by the formula C(n, r) = (n!) / [(r!) x (n – r)!] The factorials, which are the products of all positive integers equal to and less than the number you are computing, must also be computed in order to use the formula for calculating combinations. The factorial in the combination formula is indicated by the symbol (! ), which stands for the factorial functions you must use in your calculation.
What is a combination?
Combinations are the various configurations that can be produced when a sample of values or items from a larger set is taken. How many subsets you can create from the entire set of items is demonstrated by the combinations you can make. In a mathematical combination, the order of items is unimportant. This means that although some combinations produce an ordered sequence, resulting in a permutation, combinations can be formed in any order. The combinations (0, 0, 1, 1, 2, 3) and (0, 1, 2, 3, 4) are examples of non-repeating combinations.
How to calculate combinations
To determine how many combinations you can get from a sample set, use the following steps and the formula C(n, r) = (n!) / [(r!) x (n – r)!]
1. Determine your r and n values
By selecting a smaller set of items from a larger set, you can determine the values of r and n.
Assume, for instance, that you want to select four books from a shelf of eight to read. The (r) denotes the sample of the four books you select, and the (n) denotes the larger set of eight books in the formula. When you arrive at these numbers, replace the (r) and (n) variables in the formula with them:
C(n, r) = (8!) / [(4!) x (8 – 4)!]
2. Subtract your r value from your n value
Subtract these two numbers when determining your sample set and larger set for the r and n variables in the formula. Find the difference using the n = 30 books and r = 10 books from the previous step as an example:
C(n, r) = (8!) / [(4!) x (8 – 4)!] =
C(n, r) = (8!) / [(4!) x (4)!]
3. Expand on the factorials
After simplifying the formula’s expressions, you can begin computing each of the problem’s factorials. Expand each factorial in the following formula using the larger set of eight books, the sample set of four books, and the difference you discover when you subtract these values:
8! = 8 7 6 5 4 3 2 1
= 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1 = 4! x 4!
4. Cancel like terms and divide
Since these values are connected through division, you can eliminate the like terms between your factorials in the formula before dividing them. The common terms on the top and bottom of the fractional part of the equation are 4, 3, 2, 1 between 8 and (4! x 4!) Canceling these terms out results in:
C(n, r) = (8x7x6x5) / (4x3x2x1) =
C(n, r) = (1,680) / (24) =
C(n, r) = 70
When four books are selected from a set of eight, the result is 70, which is the number of combinations possible when n r 0. This means that your (n) value must be greater than or equal to your (r) value, and that all values you work with must be greater than or equal to zero. Additionally, your values must not repeat when calculating how many combinations you can get using this formula, and the order is irrelevant.
Combinations vs. permutations
Combinations can have both repeating and non-repeating arrangements. But it doesn’t matter in what order you find these arrangements. However, since permutations must occur in an ordered set, the order is crucial when dealing with them. The distinction between these two ideas can be remembered by recalling that a permutation is an ordered combination.
Permutations, like combinations, can have repeating or non-repeating values inside a set. Assume, for instance, that you need to discover a lock’s combination. The lock’s combination is an example of a permutation where the order of the numbers is crucial to the combination’s functionality. However, you might have a non-repeating permutation (such as the combination 3, 4, 5) or a repeating permutation (such as the combination 4, 4, 4).
You can use two formulas to calculate combinations when the order matters, as it does with permutations:
You can determine the permutations in the following example issue by using these formulas:
Let’s say you need to figure out the combination to a lock, and you have a choice of six numbers. Six represents your (n) value in both formulas. Three becomes your (r) value, or the smaller subset you select from the entire set, if the lock requires three numbers as part of the combination. Use the formula nr to determine how many combinations you can create using repeating values:
n^r = (6) ^ (3) =
*6 x 6 x 6 = 216
As a result, you can find 216 combinations that have repeating values. To determine the possible combinations for each of the three items you select from the set of six values, however, if the lock combination has non-repeating values, use the formula (n!) / (n – r)!
(n!) / (n – r)! =
(6!) / (6 – 3)! =
(6 x 5 x 4 x 3 x 2 x 1) / (3)! =
(3 x 2 x 1) / (6 x 5 x 4 x 3 x 2 x 1) equals
6 x 5 x 4 / 1 =
120 / 1 = 120
This finding shows that a lock with values arranged in permutations has 120 possible combinations. When calculating non-repeating permutations, you can simply reduce the factorials that are shared by each factor you divide in the formula, similar to when calculating combinations where the order is irrelevant.
How do you calculate the number of possible combinations?
We will use the formula nCr = n! / r! * (n – r)! to calculate combinations, where n denotes the total number of items and r denotes the number of items being chosen at a time. You must compute a factorial in order to calculate a combination.
How many combinations of 4 items are there?
The answer to this question (which you got right) is 24 if you meant to say “permutations.” If so, then you were likely asking “how many different ways can I arrange the order of four numbers?” Here’s how to observe this: 1.
How many combinations of 2 numbers are there?
Additionally, I am aware that there are 10 combinations of a single digit from 0–9 and 100 combinations of two digits from 0–9. I know this because 10 10 equals 100 two-digit combinations.