02 // Visual Math & Finance SolversnPr = n! / (n − r)!, nCr = n! / [r!(n − r)!]
Visual Permutations ($nPr$) vs. Combinations ($nCr$) Explorer
Selecting $r$ items from $n$ distinct elements yields $P(n,r) = \frac{n!}{(n-r)!}$ ordered permutations, and dividing out the $r!$ internal orderings yields $C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$ unordered combinations.
nPr = n! / (n − r)!, nCr = n! / [r!(n − r)!]Inputs & Parameters (2)
1±112
0±112
Combinations C(7, 3)
35
Unordered subsets (Symmetry: C(7,4) = 35)
Permutations P(7, 3)
210
Ordered arrangements = C(7,3) × 3!
Internal Orderings (3!)
6
Total n! = 5,040
Row 7 Subset Sum (2^n)
128
Binomial term: 35·a^4·b^3
20-Row Comparison Table
Compare across:
| # | Total Items in Pool (n) | Combinations C(n, r) | P(1, 1) | Row Sum 2^1 |
|---|---|---|---|---|
| 01 | 1 | 1 | 1 | 2 |
| 02 | 2 | 1 | 2 | 4 |
| 03 | 2 | 1 | 2 | 4 |
| 04 | 3 | 1 | 6 | 8 |
| 05 | 3 | 1 | 6 | 8 |
| 06 | 4 | 4 | 24 | 16 |
| 07 | 4 | 4 | 24 | 16 |
| 08 | 5 | 10 | 60 | 32 |
| 09 | 6 | 20 | 120 | 64 |
| 10 | 6 | 20 | 120 | 64 |
| 11 | 7 (Current) | 35 | 210 | 128 |
| 12 | 7 (Current) | 35 | 210 | 128 |
| 13 | 8 | 56 | 336 | 256 |
| 14 | 9 | 84 | 504 | 512 |
| 15 | 9 | 84 | 504 | 512 |
| 16 | 10 | 120 | 720 | 1,024 |
| 17 | 10 | 120 | 720 | 1,024 |
| 18 | 11 | 165 | 990 | 2,048 |
| 19 | 11 | 165 | 990 | 2,048 |
| 20 | 12 | 220 | 1,320 | 4,096 |
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Key Analytical Takeaways
- 01Permutations ($nPr$) are always $r!$ times larger than Combinations ($nCr$) because every single combination of $r$ items can be rearranged in $r!$ ordered ways.
- 02Symmetry identity: choosing $r$ items to include is identical to choosing $n - r$ items to leave behind, so $\binom{n}{r} = \binom{n}{n-r}$.
- 03The highlighted cell in row $n$, entry $r$ of Pascal's Triangle displays the exact binomial coefficient $\binom{n}{r}$.
Frequently Asked Questions
When should I use Permutations (nPr) vs. Combinations (nCr)?+
Use Permutations when order matters (podium finishes, PIN codes, seating charts). Use Combinations when order does not matter (committee selection, lottery hands, toppings).
Why is 0! (zero factorial) defined as 1?+
There is exactly 1 way to arrange an empty set of zero objects, and setting 0! = 1 ensures C(n, n) = n! / (n! × 0!) = 1.
Why does 0! (zero factorial) equal 1?+
0! = 1 represents the single empty arrangement of zero objects and preserves combinatorial identities like C(n, n) = n! / (n! · 0!) = 1.