Multinomial Coefficient Calculator — n! / (k1!·k2!·…)

Real-world use cases

• Count the distinct anagrams of a word

  A puzzle asks how many distinct ways the letters of BANANA can be arranged. Group the letters by frequency: B appears once, A three times, N twice, so enter 1, 3, 2. The tool reads n = 6 and returns 6! / (1!3!2!) = 60. The same approach handles any word with repeated letters, where a plain 6! would massively overcount because swapping two identical A's produces the same string.

• Find a coefficient in a multinomial expansion

  You need the coefficient of x^2·y·z in (x + y + z)^4 for an algebra assignment. The exponents are the group sizes, so enter 2, 1, 1. The tool reads n = 4 and returns 4! / (2!1!1!) = 12, which is exactly that coefficient. The substituted formula is shown so you can copy the full working straight into your answer rather than just the final number.

• Count ways to deal cards or split a group into teams

  Dealing 52 cards into four labelled hands of 13 each is 52! / (13!)^4. Enter 13, 13, 13, 13 and read the 29-digit answer exactly. The same calculation splits 12 students into a group of 5, a group of 4 and a group of 3 by entering 5, 4, 3, which gives 27720 distinct assignments to those labelled groups.

• Teach the link between factorials, binomial and multinomial counts

  Showing a class that C(5, 2) is just the two-group multinomial 5! / (2!3!) makes the generalisation click. Enter 2, 3 to get 10, then add a third group and watch the count change. Because the group list lives in the URL, you can prepare several worked examples as links and drop them into a lesson or a shared worksheet.

• Sanity-check a probability with the multinomial distribution

  The multinomial coefficient is the counting factor in the multinomial probability formula. Rolling a die 6 times and asking for exactly two 1's, two 2's and two 3's uses the coefficient 6! / (2!2!2!) = 90. Enter 2, 2, 2 to get that 90, then multiply by the per-outcome probability to finish the calculation by hand with confidence.

Common pitfalls

• Typing n itself as one of the inputs. This tool reads n as the sum of the group sizes, so you only enter k1, k2, …. For MISSISSIPPI you enter 1, 4, 4, 2 (the letter counts), not 11. Adding an extra 11 would push n to 22 and give a meaningless answer.

• Forgetting to group repeated items before counting arrangements. The distinct orderings of a word with repeats is n! divided by the factorial of each repeat count, not plain n!. Using 11! for MISSISSIPPI overcounts by a factor of 4!·4!·2! = 1152 because it treats identical letters as distinguishable.

• Confusing labelled groups with interchangeable ones. The multinomial coefficient counts assignments to labelled groups. If the groups themselves are not distinguishable, for example three unlabelled teams of equal size, you must divide further by the factorial of the number of equal groups to avoid overcounting.

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