Exam in Examination

Five of the letters are different while there are three alike pairs. So we have three cases:
$Case 1:$ 2 same and 2 same $\implies C\dbinom{3}{2}*\dfrac{4!}{2!2!} = 18$
$Case 2:$ 2 same and 2 different $\implies C\dbinom{3}{1}*C\dbinom{7}{2}*\dfrac{4!}{2!}= 756$
$Case 3:$ All different $\implies C\dbinom{8}{4}*4! = 1680$
So, the number of permutations is $18+756+1680=\boxed{2454}$

We can consider the multiset $(1.E,1.X,2.A,1.M,2.I,2.N,1.T,1.O)$

Hence the required exponential generating function is

$G(x)=(1+\frac{x}{1!})^{5}(1+\frac{x}{1!}+\frac{x^{2}}{2!})^{3}$

In the above, the co-efficient of $\frac{x^{4}}{4!}$ is $2454$ .

I basically used the following theorem:

The number of $n$ -permutations of the multiset $(n_{1}.a_{1},...,n_{k}.a_{k})$

is given by the co-efficient of $\frac{x^{n}}{n!}$ in

$G(x)=f_{n_{1}}(x)...f_{n_{k}}(x)$ where $f_{n_{i}}(x)=1+x+\frac{x^{2}}{2!}+..+\frac{x^{n_{i}}}{n_{i}!}$

The word examination consists of 11 letters -

(AA), (II), (NN), E, X, M, T, O.

The following combinations are possible:

(a) 2 alike, 2 alike:

³C2 = 3 ways

(b) 2 alike, 2 different:

³C1×7C2 = 63 ways

(c) all 4 different:

8C4 = 70 ways

Hence required number of combinations = 3 +63 +70 = 136.

To find the number of permutations: In (a), the number of permutations = 3×4!/[2!2!] = 18

In (b), the number of permutations = 63×4!/2! = 756

In (c), the number of permutations = 70×4!= 1680

Hence required number of permutations = 18 +756 +1680 = 2454.