How many permutations of 4 letters can be made out of the letters of the word 'examination'?

Question

[as stated in the title]

[n-p-n-p-n-p]

Answer 1

The word examination has $11$ letters. It is composed of $8$ different letters, $3$ of which repeat once.

As Andre suggests, break it down into three cases.

1. All different. Choose the four letters out of the eight you'll use ${8 \choose 4}$, and then order them ($4!$).
2. One double. Choose which of the three letters is the double ${3 \choose 1}$, then choose which positions they go in the word ${4 \choose 2}$. Then, choose the two remaining letters ${7 \choose 2}$, and decide which goes first and which goes second in the remaining two spaces ($2!$).
3. Two doubles. Choose which two of the three doubled letters to use ${3 \choose 2}$, then order them ($6$).

Adding all of these gives $2454$.

Answer 2

We could also use generating functions (which has recently been my favorite method since I suck at identifying cases).

Consider the multiset $M=[1\text{E}, 1\text{X}, 2\text{A}, 1\text{M}, 2\text{I}, 2\text{N}, 1\text{T}, 1\text{O}]$.

The number of $4$-permutations of $M$ is the coefficient of $x^4/4!$ in the exponential expansion of

$$(1+x)^5 \left(1+x+\dfrac{x^2}{2!}\right)^3.$$

The coefficient of $x^4$ above is $409/4$. Multiplying by $4!$ we get $2454$.

---

The $(1+x)$ represent the letters that are not repeated, while $(1+x+x^2/2)$ represents those letters that are repeated twice.