How many 3 letter words with or without meaning can be formed out of the letters of the word algebra if repetition of letters is not allowed?
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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.
Answer: Option B Explanation: The word 'LOGARITHMS' has 10 different letters. Hence, the number of 3-letter words(with or without meaning) formed by using these letters zulqarnain 2014-12-04 16:52:03 why we used permutation here instead of combination? Raju 2014-12-11 19:59:44 because order matters. for example, LOG and GOL are different words.  sara 2015-09-29 11:42:48 but how can we detect that the problem require a permutation, it is not even mentioned that order matter Sam 2015-09-29 13:30:51 Question is to find out the number of 3-letter words that can be formed out of the letters of the LOG and GOL are two different words. Therefore order matters. Hence, permutation. Add Your Comment(use Q&A for new questions) Name The correct option is A 720 Step 1: Use combination formula In the word LOGARITHMS there are 10 unique letters which are, A,G,H,I,L,M,O,R,Sand T. Now we must create a three-letter word with or without meaning, with the restriction that letter repetition is not permitted, i.e., we cannot use the same letter more than once to create three-letter words. We know that number of combinations of r objects chosen from n objects when repetition is not allowed is given by Crn=n!r!(n-r)! where n! is n!=n×(n–1)×(n–2)×(n–3)×……..×3×2×1 So, three letters out of 10 unique letters can be selected in C310ways. By using the above formula we get C 310=10!3!(10-3)! =10!3!(7) ! Step 2: Calculate the number of 3-letter words In general, n! can be used to arrange n distinct objects. We chose three letters from a list of ten unique letters, and these letters can be put in three different ways. Total number of 3 letter word =C310×3! ∴ C310×3!=10!3!(7)!×3! =10 !7! =10×9×8×7!7! =10×9 ×8 =720 Hence, the word LOGARITHMS if repetition of letters is not allowed can form 720 number of 3-letter words. |
