Permutations Calculator (nPr)
Enter the number of elements (n) and the number of the sample (r), \(P(n,r)\).
Answer:
Step-by-step solution:
Use this calculator to find the number of permutations possible by choosing a certain number of elements from a set. This calculator finds the number of subsets that can be obtained when the order of the elements does matter. Enter the values of n and r in the corresponding input boxes.
How to use the permutation calculator?
Step 1: Enter the total number of items in the first input box. This is the value of n.
Step 2: Enter the sample number in the second input box. This is the value of r.
Step 3: Click “Calculate” to get the number of permutations that corresponds to the values entered.
Step 4: The answer will be displayed on the right-hand side and the step-by-step solution will be displayed on the bottom.
What types of numbers can I use in the calculator?
To calculate permutations we must only consider positive integers. This is because we associate permutations with elements of a set and we cannot have a fraction of an element. We must have integer elements.
Additionally, the sample number must be less than or equal to the total number of items. This means that the numbers entered must meet the following condition:
\(n\geq r\geq 0\)
What are permutations?
A permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the elements matters. Many times they are confused with combinations, but the difference is that, in combinations, the order of the elements does not matter and in permutations it does.
To learn more about permutations, take a look at our article Permutations – Example and Practice Problems.
How to calculate permutations?
To find permutations, we can use the permutations’ formula:
\( P(n,~r)=\frac{n!}{(n-r)!}\)
Where n is the total number of elements in a set and r is the number of selected elements.
For example, suppose we want to calculate the number of possible permutations by having a set of 6 elements and choosing 3. Then, we have:
\( P(n,~r)=\frac{n!}{(n-r)!}\)
\( P(6,~4)=\frac{6!}{(6-3)!}\)
\(=\frac{6!}{(3)!}\)
We can rewrite 6! like 6×5×4×3! Then, we can simplify the numerator with the 3! of the denominator. Therefore, we have:
\(P(6,~4)=\frac{6!}{(3)!}\)
\(=\frac{6\times 5 \times 4 \times 3!}{(3)!}\)
\(=6\times 5 \times 4\)
\(=120\)
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