Combinations (nCr)

Builds onPermutations (nPr) — if this page feels steep, start there.

{}_{n}C_{r} = \binom{n}{r} = \frac{n!}{(n-r)!\,r!}

${}_{n}C_{r}$: read 'n choose r': the number of different groups of r you can pick from n
$\binom{n}{r}$: the same thing written as a stacked pair — just alternative notation
$n!$: the exclamation mark is 'factorial': n × (n−1) × (n−2) … × 1
$(n-r)!\,r!$: the denominator removes the orderings we don't care about — r! ways to shuffle each chosen group

Why it must be true

How many ways can you choose $r$ things from $n$ when order doesn't matter — say, picking 4 stocks for a portfolio from a 10-stock watchlist? Start with ordered arrangements ( $n!/(n-r)!$ ), then notice every chosen group has been counted once for each of its $r!$ internal orderings. Divide that redundancy out.

The test for "combination vs. permutation" is one question: would swapping two selections produce a different outcome? A portfolio doesn't care which stock you picked first — divide by $r!$ .

The derivation

Count ordered selections first: $n$ choices, then $n-1$ , … for $r$ picks:

{}_{n}P_{r} = \frac{n!}{(n-r)!}

Each unordered group of $r$ items appears $r!$ times in that count — once per internal ordering. Removing the duplication:

\binom{n}{r} = \frac{{}_{n}P_{r}}{r!} = \frac{n!}{(n-r)!\,r!}

When to reach for it

Counting how many distinct GROUPS can be selected when the order of selection is irrelevant.

Listen for

“how many ways to choose”“committee / basket / portfolio of r from n”“order does not matter”

Back-of-the-envelope

Estimate it in your head first — then the calculator only confirms.

≈
Use the small side: $\binom{n}{r} = \binom{n}{n-r}$ , so 10 choose 8 is just 10 choose 2 = 45.
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n choose 2 is the handshake number: n(n−1)/2. Memorize 10C2 = 45, 10C3 = 120 as anchors.
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Combinations < permutations, always. If both counts appear as choices, the smaller of the pair is the combination.

Traps in applying it

✗Counting ordered arrangements (permutations) when swapping picks changes nothing.
✗Allowing the same item to be picked twice — selection here is without replacement.
✗Answering for r − 1 or n − 1 after misreading the pool.

Limits & criticisms

The formula presumes distinguishable items chosen without replacement — repetition or identical items need different counts. And factorials explode: beyond small n the raw counts overflow intuition (and calculators), which is why real applications lean on logarithms or approximations.

Where it came from

Counting selections is genuinely ancient: the Indian prosodist Pingala (~200 BCE) counted poetic meters, and the array we call Pascal's triangle was known in India, Persia and China centuries before Pascal's 1654 treatise connected it to probability. Jacob Bernoulli's Ars Conjectandi (1713) made combinatorics the grammar of chance.

Today the binomial coefficient sits inside the binomial option-pricing model (counting up-down paths), credit portfolio math, and any question of the form "how many distinct baskets can I build?"

Drill this formula Back to the library