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Binomial Coefficients

Binomial Coefficients

Numbers written in any of the ways shown below. Each notation is read aloud "n choose r."

Three equivalent notations for binomial coefficients: (n choose r), nCr, C(n,r), and C with superscript n and subscript r.

A binomial coefficient equals the number of combinations of r items that can be selected from a set of n items. It also represents an entry in Pascal's triangle. These numbers are called binomial coefficients because they are coefficients in the binomial theorem.

 

Formula: Formula: C(n,r) or nCr = n! / (r!(n-r)!) = n(n-1)(n-2)···(n-r+1) / r!
  Note: Formula showing binomial coefficient: (n choose r) equals n P r divided by r factorial, where nPr is the formula for permutations of n objects taken r at a time.
Examples:

7 choose 2 equals 7! divided by 2!5!, equals (7 times 6) divided by (2 times 1), equals 21

 


Formula: (7 choose 5) = 7!/5!2! = (7·6·5·4·3)/(5·4·3·2·1) = (7·6)/(2·1) = 21

20C3 = 20!/(3!17!) = (20·19·18)/(3·2·1) = 1140

 

See also

Factorial, binomial coefficients in Pascal's triangle, combination formula, permutations

Key Formula

(nr)=n!r!(nr)!\binom{n}{r} = \frac{n!}{r!\,(n-r)!}
Where:
  • nn = Total number of items in the set (a non-negative integer)
  • rr = Number of items being chosen, where 0 ≤ r ≤ n
  • !! = Factorial — the product of all positive integers up to that number (e.g., 5! = 120)

Worked Example

Problem: Compute the binomial coefficient (83)\binom{8}{3}.
Step 1: Write out the formula with n = 8 and r = 3.
(83)=8!3!(83)!=8!3!5!\binom{8}{3} = \frac{8!}{3!\,(8-3)!} = \frac{8!}{3!\cdot 5!}
Step 2: Expand only the numerator factors that don't cancel with 5! in the denominator. Since 8! = 8 × 7 × 6 × 5!, the 5! cancels.
8!3!5!=8×7×63!\frac{8!}{3!\cdot 5!} = \frac{8 \times 7 \times 6}{3!}
Step 3: Evaluate 3! in the denominator.
3!=3×2×1=63! = 3 \times 2 \times 1 = 6
Step 4: Divide to get the final result.
8×7×66=3366=56\frac{8 \times 7 \times 6}{6} = \frac{336}{6} = 56
Answer: (83)=56\binom{8}{3} = 56. There are 56 ways to choose 3 items from a set of 8.

Another Example

This example applies the binomial coefficient to a real-world counting problem, reinforcing that 'n choose r' counts combinations. It also uses larger numbers than the first example, giving practice with the cancellation shortcut.

Problem: A pizza shop offers 10 toppings. You want to order a pizza with exactly 4 toppings. How many different pizzas can you create?
Step 1: Identify n and r. You have n = 10 toppings and are choosing r = 4.
(104)=10!4!6!\binom{10}{4} = \frac{10!}{4!\cdot 6!}
Step 2: Cancel 6! from the numerator and denominator. This leaves the top three factors of 10! above 6.
10!4!6!=10×9×8×74!\frac{10!}{4!\cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4!}
Step 3: Compute 4! = 24 and multiply the numerator.
504024=210\frac{5040}{24} = 210
Answer: You can create 210 different 4-topping pizzas from 10 available toppings.

Frequently Asked Questions

What is the difference between a binomial coefficient and a permutation?
A binomial coefficient (nr)\binom{n}{r} counts the number of unordered selections (combinations) of rr items from nn. A permutation P(n,r)=n!(nr)!P(n, r) = \frac{n!}{(n-r)!} counts ordered arrangements. Since order doesn't matter in a combination, (nr)\binom{n}{r} is always less than or equal to P(n,r)P(n, r); specifically, (nr)=P(n,r)r!\binom{n}{r} = \frac{P(n,r)}{r!}.
Why is any number 'choose 0' equal to 1?
By the formula, (n0)=n!0!n!=1\binom{n}{0} = \frac{n!}{0!\cdot n!} = 1 because 0!=10! = 1 by definition. Intuitively, there is exactly one way to choose nothing from a set: you simply select no items at all.
How are binomial coefficients related to Pascal's triangle?
Each entry in Pascal's triangle is a binomial coefficient. The entry in row nn and position rr (both starting from 0) equals (nr)\binom{n}{r}. The familiar addition rule of Pascal's triangle corresponds to the identity (nr)=(n1r1)+(n1r)\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}.

Binomial Coefficient (Combination) vs. Permutation

Binomial Coefficient (Combination)Permutation
DefinitionNumber of unordered selections of r items from nNumber of ordered arrangements of r items from n
Formulan!r!(nr)!\frac{n!}{r!(n-r)!}n!(nr)!\frac{n!}{(n-r)!}
Order matters?No — {A, B} and {B, A} count as the same selectionYes — AB and BA count as different arrangements
Relationship(nr)=P(n,r)r!\binom{n}{r} = \frac{P(n,r)}{r!}P(n,r)=r!(nr)P(n,r) = r! \cdot \binom{n}{r}
Example (n=5, r=3)(53)=10\binom{5}{3} = 10P(5,3)=60P(5,3) = 60

Why It Matters

Binomial coefficients appear throughout algebra, probability, and statistics. You use them whenever you need to count combinations—for instance, choosing a committee from a group, finding probabilities in binomial distributions, or expanding expressions like (x+y)n(x + y)^n. Mastering them is essential for courses from Algebra 2 through AP Statistics and beyond.

Common Mistakes

Mistake: Confusing the combination formula with the permutation formula by forgetting the r!r! in the denominator.
Correction: The combination formula divides by both r!r! and (nr)!(n-r)!, while the permutation formula divides only by (nr)!(n-r)!. Always check whether order matters: if it does not, you need the extra r!r! in the denominator.
Mistake: Computing (nr)\binom{n}{r} by fully expanding n!n! for large nn, leading to enormous numbers and arithmetic errors.
Correction: Cancel common factors first. For (nr)\binom{n}{r}, write only the rr factors n×(n1)××(nr+1)n \times (n-1) \times \cdots \times (n-r+1) in the numerator and divide by r!r!. This keeps the numbers manageable.

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