GATE CS 2016 Set 1 Q.36 (CS Q.26): Combinatorics, Counting, Recurrence Relations and Generating Functions

The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + \cdots)^3$ is _____.

Your answer

Official answer and solution

Solution

Factor out $x^9$ and apply the negative binomial series expansion.

We seek the coefficient of $x^{12}$ in:

P(x) = (x^3 + x^4 + x^5 + x^6 + \cdots)^3

Factor out $x^3$ from the infinite sum:

x^3 + x^4 + x^5 + \cdots = x^3(1 + x + x^2 + \cdots) = \frac{x^3}{1 - x}

Cubing this expression gives:

P(x) = \left(\frac{x^3}{1 - x}\right)^3 = x^9 (1 - x)^{-3}

Finding the coefficient of $x^{12}$ in $x^9 (1 - x)^{-3}$ is equivalent to finding the coefficient of $x^{12 - 9} = x^3$ in $(1 - x)^{-3}$ .

Recall the negative binomial series expansion:

(1 - x)^{-n} = \sum_{r=0}^\infty \binom{n + r - 1}{r} x^r

For $n = 3$ and $r = 3$ :

\text{Coefficient} = \binom{3 + 3 - 1}{3} = \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10

Final answer

Common trap

Expanding $(1 - x)^{-3}$ directly for $r = 12$ without factoring out $x^9$ .

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