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.
9. How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed? | |
A. 5040 | B. 720 |
C. 420 | D. None of these |
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
= 10P3
$=10×9×8\\=720$
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
word 'LOGARITHMS'
LOG and GOL are two different words. Therefore order matters. Hence, permutation.
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[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.