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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.

TESLA____
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  • No of distinct letters are S,E, R,I and gives a total of three letter words that could be formed with or without meaning would be $3^4$ where each of these letters could be repeated. – Satish Ramanathan Oct 30 '14 at 19:15
  • I think you need to split this into cases depending on which letters you select. – Daniel Goldman Oct 30 '14 at 19:15

4 Answers4

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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.

N. F. Taussig
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You have 4 distinct letters, where 2 of those can be used twice. In a three letter word, you can have 0 repetition, or 1 repetition. Number of three letter words with zero repetition: $4! = 24$, and number of three letter words with one repetition is $\binom21\binom323 = 18$, so total is $24+18=42$

Rey
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Their are six letters. Out of which 2e,2s, r and I. For permutations with no repetition- permutation of letters (s, e, r, I) taking 3 at a time = 4p3=24. For repetition of e and s- suppose we take (ss ) now third letter can be chosen out of e, r and I in 3c1=3 ways. Now each of this can be arranged in 3!/2!=3 therefore total number of permutation with repeting s is 3c1×(3!/2!)=9. Same is the case with repeting letter e. Therefore total permutation with repeting s and e is 9×2=18. Thus total permutation of the letters of the word series is 24+18=42.

Akhil Wadichar
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SERIES has the repeated letters (SS) und (EE), making 4 distinctive letters.

  1. A three letter word when all the letters are different, the number of the ways is $C\binom{4}{3} \cdot 3! = 24$ ways.

  2. 2 letters are same, 1 letter is different:

Number of the ways is $C\binom{2}{1} \cdot C\binom{3}{1} \cdot \frac{3!}{2!} = 18$ ways.

Total is 42.

peterh
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Mohamed K Musoke
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