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How many terms are in a binomial expansion?
If you expand (a + b)ⁿ, the power of a starts at n and decreases down to 0, while the power of b starts at 0 and increases to n. Because the power counts from 0 up to n, there are exactly n + 1 terms in total.
For example, (a + b)² = a² + 2ab + b², which has 2 + 1 = 3 terms. When working on more complex problems, you can read more about Binomial theorem: general term, middle term and coefficients.
What is the number of terms in a plus b plus c power n?
For a trinomial expansion (a + b + c)ⁿ, the counting becomes a combinatorics problem. Every term in the expansion contains a product of the three variables with non-negative integer powers that must add up exactly to n (for example, powers x, y, and z where x + y + z = n).
The total number of such terms is the number of ways to distribute the total power n across the three variables. Using the stars and bars method, the formula for the number of terms is (n + 1)(n + 2) / 2 (Wikipedia).
This arrangement forms a triangular pattern. Just as binomial terms sit on a line, trinomial terms sit on a triangle, which is why the answer is always a triangular number. If this combinatorial counting feels like a stretch, the Fermi Concept Library has every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson and its place in the map.
How do you find the terms for products like (1 + x)ⁿ(1 − x)ⁿ?
Examiners in exams like JEE Main often test if you blindly apply formulas or look at the structure first. Consider the algebraic product (1 + x)ⁿ(1 − x)ⁿ.
Instead of treating this as two separate expansions to multiply out, you can combine the bases using the difference of squares:
(1 + x)ⁿ(1 − x)ⁿ = ((1 + x)(1 − x))ⁿ = (1 − x²)ⁿ
By simplifying first, the expression collapses back into a standard binomial raised to the power of n. Therefore, it will simply have n + 1 terms, not a larger number derived from expanding and multiplying.
How do you count terms when expansions are added together?
Sometimes, you need to combine the binomial and trinomial rules to solve a single problem. Consider finding the total number of terms in the expansion of (a + b)³ + (x + y + z)².
Evaluate the binomial part: (a + b)³ has n = 3, so it produces 3 + 1 = 4 terms.
Evaluate the trinomial part: (x + y + z)² has n = 2. Using the formula (2 + 1)(2 + 2) / 2, we find it produces 3 × 4 / 2 = 6 terms.
Check for overlapping variables: Since the variables (a, b) and (x, y, z) are entirely different, there are no like terms that can be added together to simplify the expression.
Add the counts: The total number of terms is simply 4 + 6 = 10 distinct terms.
What if the variables overlap? If you expand (a + b)² + (a - b)², you must expand and collect like terms. The first part gives a² + 2ab + b² and the second gives a² - 2ab + b². The 2ab and -2ab cancel out, leaving 2a² + 2b², which is only 2 distinct terms. Always look at the structure and overlapping variables before mechanically applying the counting rules.
What is the general rule for multinomials?
What if you have four terms, like (a + b + c + d)ⁿ? The same stars and bars logic applies to any number of variables. For k variables raised to the power of n, the number of terms is given by the combination C(n + k − 1, k − 1).
For k = 4 and n = 3, the number of terms is C(3 + 4 − 1, 4 − 1), which equals C(6, 3), or 20. This shows the rapid growth of algebraic terms, a concept closely tied to Permutations and combinations: how to stop double counting. Also, the properties of these terms relate back to Sums of binomial coefficients: C₀ + C₁ + … and alternating sums.
If you keep double counting or missing terms in expansion problems, the Fermi Problem Library offers worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
