Problem Statement

How many terms are there in the expansion of

\[(x_1 + x_2 + x_3 + x_4 + x_5 + x_6)^{18}\]

after all like terms have been combined?

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Solution

Step 1: Understanding the Expansion

When expanding

\[(x_1 + x_2 + x_3 + x_4 + x_5 + x_6)^{18}\]

each term in the expansion will be of the form:

\[c \cdot x_1^{a_1} x_2^{a_2} x_3^{a_3} x_4^{a_4} x_5^{a_5} x_6^{a_6}\]

where \(a_1, a_2, a_3, a_4, a_5, a_6\) are non-negative integers that sum to 18:

\[a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 18.\]

Each unique term corresponds to a different way of assigning exponents to the six variables, regardless of order.

Step 2: Reduced Version - The Core Idea

Instead of thinking in terms of exponents, simply answer:

“In how many ways can I divide 18 objects amongst 6 people, if each person can get 0 at min?”

This is a direct application of the Stars and Bars method, which states that the number of ways to distribute \(n\) identical objects among \(k\) groups is:

\[\binom{n + k - 1}{k - 1}.\]

Setting \(n = 18\) (total power) and \(k = 6\) (number of variables), we get:

\[\binom{18 + 6 - 1}{6 - 1} = \binom{23}{5}.\]

Step 3: Compute the Binomial Coefficient

Using the formula:

\[\binom{23}{5} = \frac{23!}{5!(23 - 5)!} = \frac{23!}{5! \cdot 18!}.\]

Computing step-by-step:

\[\frac{23 \times 22 \times 21 \times 20 \times 19}{5 \times 4 \times 3 \times 2 \times 1}\] \[= \frac{5173168}{120} = 33649.\]

Thus, the number of terms in the expansion is \(\mathbf{33649}\)

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