How many three letter words without meaning are possible when repetition of letters is not allowed?

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.

Text Solution

Answer : (i) 60 (ii) 125

Solution : (i) Total number of 3-letter words is equal to the number of ways of filling 3 places. First place can be filled in 5 ways by any of the given five letters. Second place can be filled in 4 ways by any of the remaining 4 letters and the third place can be filled in 3 ways by any of the remaining 3 letters. <br> So, the required number of 3-letter words `=(5xx4xx3)=60.` <br> (ii) When repetition of letters is allowed, each place can be filled by any of the 5 letters in 5 ways. <br> `therefore " the required number of ways "=(5xx5xx5)=125.`

How many 3 letter words with or without meaning can be formed out of the letters of the word SMOKE when repetition of words is allowed?

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