Mathematics · pattern A, Numeric fields
Enter n and r for the number of selections, with the permutation count and the factorials behind both shown as working.
ⁿCᵣ = n! ÷ (r!(n − r)!)
Combinations are permutations divided by r!, which removes the orderings within each selection. They are the binomial coefficients.
13,983,816
n = 49, r = 6. Permutations count arrangements where order matters; combinations count selections where it does not.
1ⁿPᵣ = n! ÷ (n − r)! = 608,281,864,034,267,500,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 ÷ 60,415,263,063,373,830,000,000,000,000,000,000,000,000,000,000,000,000 = 10,068,347,520
arrangements of r items chosen from n, order significant
2ⁿCᵣ = ⁿPᵣ ÷ r! = 10,068,347,520 ÷ 720 = 13,983,816
divide out the r! orderings of each selection
ⁿCᵣ are exactly the entries of Pascal's triangle and the coefficients of the binomial expansion, which is why the same values appear in probability, algebra and combinatorics. The symmetry ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ is also useful: choosing 45 from 49 is the same count as choosing 4.
If the number is not the part you are stuck on, that is what the service is for — a specialist who explains the working, not just the answer.
Arithmetic runs in double-precision floating point, so results beyond about fifteen significant figures are not exact. Where a question wants an exact fraction or surd, keep the exact form rather than a decimal.