Words of length 3 specific letters. If I have the letters M A T E M A T I K I, how many words of length 3 can I build from these?
My thoughts are that for length 3, I have $6!$ words of only single letters, like MAT, MAE, IKT and so on. Words with double letters I have $4*5*{3 \choose 2}$. So the total number of words of length 3 would be $6! + 60$. Is this correct?
 A: In case the word can only contain distinct letters, we have just 6 letters to choose from, so we have $6 \choose 3$ different combinations, and because we need them ordered, then there are $6 \choose 3$ x $3!$ possibilities.
In case the word can contain doubles, then we have 4 cases, and each case has $5 \choose 1$ x $3$ possibilities. In total, we have $4$ x $3$ x $5 \choose 1$ $=$ $60$ possibilities.
Example: If the two (identical) letters to be chosen are (MM), then the possibilities are: with I, (MMI) (MIM) (IMM), etc.
Add the two results together.
A: MATEMATIKI contains 4 letters twice, and two letters once, so if $a_n$ denotes the number of $n$-letter words that can be made, we have the exponential generating function representation
$$\sum_n a_n \frac{x^n}{n!} = \left(1+x+\frac{x^2}{2!}\right)^4 (1+x)^2.$$
The coefficient of $x^3$ on the right is $30$; multiplying by $3!$ gives the answer as $180$.
A: This is just an elaboration of "Ahmed Hussein's" answer.
Divide the letters as 'MM', 'AA', 'TT', 'II', 'EK'.
Now, for the first case, consider only $6$ letters: M,A,T,I,E,K. Number of words of length $3$ from these $6$ letters is $^6P_3=120$.
Next, choose any one group from the first four groups. That can be done in $^4C_1$ ways. Lets say we chose the group 'MM'. So, our word contains these two letters with place for the third letter. That has to be chosen from 'A,T,I,E,K'. So, the number of ways in which we can choose one letter out of these five is $^5C_1$. The number of arrangements of $3$ letters in which $2$ are same and $1$ is different is $\frac{3!}{2!}$. So, the total number of words you get from this case is $^4C_1 *^5C_1*\frac{3!}{2!}=60$
Hence, you have the total number of words $120+60=180$
