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Combination Calculator nCr – Choose Formula

Quick answer

C(n,r)=n!/(r!(n−r)!) counts unordered picks: C(5,2)=10 and lottery C(49,6)=13983816.

  • Compare against permutations nPr instantly.

Figures are estimates from this calculator’s standard formula — adjust the inputs in the tool below for your exact number.

How the Combination Calculator Works

C(n,r) = n! / (r! × (n−r)!)

Combinations count selections where order does not matter. Choosing 2 students from 5 gives C(5,2) = 5!/(2!×3!) = 120/12 = 10 pairs. The factorial expansion cancels nicely for hand work: C(10,3) = (10×9×8)/(3×2×1) = 720/6 = 120. Contrast with permutations, where order matters: P(5,2) = 20 because each pair {A,B} splits into AB and BA — the ratio is exactly r!, here 2. The calculator prints the canceled fraction, the decimal value and the nPr comparison together. Lottery odds make the scale vivid: C(49,6) = 13,983,816 distinct tickets, which is why a single line has about a 1-in-14-million jackpot chance. Values update live as you adjust n and r, with symmetry C(n,r) = C(n,n−r) applied to keep arithmetic small.

  1. 1

    Enter n and r

    Set the pool size n and how many you choose, r ≤ n

  2. 2

    Expand the factorials

    The tool shows n! over r!(n−r)! with the canceled shortcut

  3. 3

    Compute C(n,r)

    Get the exact count of unordered selections

  4. 4

    Compare with nPr

    See P(n,r) = C(n,r) × r! side by side

Use Cases

Lottery Odds

C(49,6) = 13,983,816 — the true ticket space of a 6/49 draw

Team Selection

Pick a 3-person squad from 10 candidates: C(10,3) = 120 options

Pizza Toppings

Choose 3 of 9 toppings: C(9,3) = 84 distinct pizzas

Tips

  • 1

    C(n,r) = C(n,n−r): choosing 2 winners equals choosing 3 losers from 5

  • 2

    Order matters? Use nPr. Order irrelevant? Use nCr

  • 3

    The canceled form (n×…×(n−r+1))/r! avoids giant factorials by hand

Common Mistakes

  • Using permutations when order is irrelevant, inflating C(5,2)=10 to 20

  • Forgetting r! in the denominator and getting 60 for C(5,2)

  • Mixing up n and r — C(2,5) is impossible since r cannot exceed n

  • Double counting subsets when generating them by hand

FAQs

How many ways can I choose 2 from 5?

C(5,2) = 5!/(2!3!) = 10. Order is irrelevant, so AB and BA are one pair.

What are my 6/49 lottery odds?

C(49,6) = 13,983,816 combinations, so one ticket wins about 1 in 13.98 million.

What is the difference between nCr and nPr?

nPr counts ordered arrangements, nCr unordered. P(5,2) = 20 = 2! × C(5,2) = 2 × 10.

What is C(10,3)?

120: (10×9×8)/(3×2×1) = 720/6.

Why is C(n,0) = 1?

There is exactly one way to choose nothing — the empty set — so C(n,0) = 1 for every n.