Oxford Entrance Exam Problem 4

Probability Level 3

$\large (1+xy+y^2)^n$

Let $n$ be a positive integer. The coefficient of $x^3y^5$ in the expansion of the expression above equals to?

\binom{n}{3}\binom{n}{5}

2^n

4\binom{n}{4}

\binom{n}{8}

n

1 solution

Raj Magesh
May 30, 2015

The general term of a trinomial expansion $(x+y+z)^n$ is:

$\left(\dfrac{n!}{a!b!c!}\right) x^a y^b z^c$

where $a+b+c=n$ .

We have $(1+xy+y^2)^n$ , whose general term is:

$\left(\dfrac{n!}{a!b!c!}\right)(1)^a (x)^b (y)^{b+2c}$

Comparing coefficients, we get $b = 3$ , $c=1$ and $a=n-4$ , making the trinomial coefficient

$\dfrac{n!}{3!(n-4)!} = 4\dbinom{n}{4}$