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Exercise 1543Combinatorics › Binomial coefficientsL1Added 3 Nov 2026

The multinomial theorem

#multinomial#expansion#coefficients

Statement

Expand (a+b+c)3(a+b+c)^3 using the multinomial theorem.

Hint

(a+b+c)3=∑i+j+k=33!i!j!k!aibjck(a+b+c)^3=\sum_{i+j+k=3}\frac{3!}{i!j!k!}a^ib^jc^k.

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Solution

Terms. The multi-indices (i,j,k)(i,j,k) with sum 33: (3,0,0)(3,0,0), (0,3,0)(0,3,0), (0,0,3)(0,0,3), (2,1,0)(2,1,0), (2,0,1)(2,0,1), (1,2,0)(1,2,0), (0,2,1)(0,2,1), (1,0,2)(1,0,2), (0,1,2)(0,1,2), (1,1,1)(1,1,1).

Coefficients. 3!i!j!k!\frac{3!}{i!j!k!}: 11 for (3,0,0)(3,0,0), 33 for permutations of (2,1,0)(2,1,0), 66 for (1,1,1)(1,1,1).

Result. (a+b+c)3=a3+b3+c3+3a2b+3a2c+3ab2+3b2c+3ac2+3bc2+6abc(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc.

Conclusion. The multinomial expansion.