To find the number of three letter words that can be formed from the word 'SERIES', with or without meaning and without repetition.
The number of permutations if all letters were distinct = $^{6}P_3$.
As 'S' and 'E' are repeating, the arrangement in which 'S' are interchanged are the same, so we divide by $2!$ for 'S' and likewise for 'E'.
Number of permutations $= \frac{^{6}P_3}{2!.2!} = 30$
Which isn't the correct answer. I would like to get some help here.
Best Answer
There are four distinct letters in the word SERIES. You can either use three of the four letters or use two of the four letters by using either S or E twice.
You can use three distinct letters in $P(4, 3) = 4 \cdot 3 \cdot 2 = 24$ ways.
You can use exactly two letters if you use S or E twice. Thus, there are $C(2, 1)$ ways of choosing the repeated letter, $C(3, 2)$ ways of choosing where to place those letters in the three letter word, and $C(3, 1)$ of choosing the third letter in the word, giving
$$\binom{2}{1}\binom{3}{2}\binom{3}{1} = 2 \cdot 3 \cdot 3 = 18$$
ways to form a word with a repeated letter.
Consequently, there are $24 + 18 = 42$ distinguishable three letter words that can be formed with the letters of the word SERIES.