How many four-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] 40
B] 400
C] 5040
D] 2520
Answer
Hint: We can take the letters in the given word and count them. Then we can find the permutation of forming 4 letters words with the letters of the given words by calculating the permutation of selecting 4 objects from n objects without replacement, where n is the number of letters in the given word which is obtained by the formula, ${}^n{P_r} = \dfrac{{n!}}{{\left[ {n - r} \right]!}}$
Complete step by step solution:
We have the word ‘LOGARITHMS’.
We can count
the letters. After counting, we can say that there are 10 letters in the given word.
$ \Rightarrow n = 10$
Now we need to form four letter words from these 10 numbers. As the words can be with or without meaning, we can take all the possible ways of arrangements.
As the repetition is not allowed, we can use the equation ${}^n{P_r} = \dfrac{{n!}}{{\left[ {n - r} \right]!}}$ where n is the number of objects and r is the number of objects needed to be selected.
So, the number of
four-letter words can be formed is given by,
$ \Rightarrow {}^{10}{P_4} = \dfrac{{10!}}{{\left[ {10 - 4} \right]!}}$
So we have,
$ \Rightarrow {}^{10}{P_4} = \dfrac{{10!}}{{6!}}$
Using properties of factorial, we can write the numerator as,
$ \Rightarrow {}^{10}{P_4} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6!}}{{6!}}$
On cancelling common terms we get,
$ \Rightarrow {}^{10}{P_4} = 10 \times 9 \times 8 \times 7$
Hence we have,
$ \Rightarrow {}^{10}{P_4} =
5040$
Therefore, the number of four-letter words that can be formed is 5040.
So the correct answer is option C.
Note: Alternate method to solve this problem is by,
We have 10 letters that have to be arranged in four places. It is given that repetition of letters is not allowed. So the letter once used cannot be used again.
So, in the $1^{\text{st}}$ place, we place any of the 10 letters. In the second place we can put any of the remaining 9 letters. In
$3^{\text{rd}}$ place we can have any of the 8 letters and in the last place any of the remaining 7 letters can be used.
So, the total arrangement is given by, $10 \times 9 \times 8 \times 7 = 5040$ .
Therefore, the number of words that can be formed is 5040.
Nội dung chính
- How many 3 letter words can be formed using a, b, c, d, e ifA.repetition of letters is not allowed?B.repetition of letters allowed?
- How many 3 letter words can be formed from signature if repetition is not allowed?
- How many 3 letter words can be made if letters can be repeated?
- How many ways 4 letters can be formed from a b/c d/e and f if repetition is allowed?
- How many 3 letter words can be formed from the letters Abcde if letters can be repeated in a word?
How many 3 letter words can be formed using a, b, c, d, e ifA.repetition of letters is not allowed?B.repetition of letters allowed?
Answer
Verified
Hint: In this question, we have to find the number of 3 letter words that can be formed using a, b, c, d, and e. If repetition is allowed then we can take that letter again. So, we have 5 intakes
for every three places of 3 letter words.
The total number of words \[=5\times 5\times 5=125\] . Now, if repetition is not allowed then we have to take only three letters out of 5 letters. The number of ways to select 3 letters out of 5 letters is \[^{5}{{C}_{3}}\] . But the letter can also be rearranged. The number of ways to rearrange 3 letters are \[3!\] . The total number of 3 letter words is \[^{5}{{C}_{3}}\times 3!\] .
Now, we have to find the number of 3 letter words if repetition is not allowed. That is we have to take 3 letters and we are given 5 letters. The number of 3 letter words that can be formed using 5 letters is \[^{5}{{C}_{3}}\] .
We know that words can be formed after rearranging also. So, the number of rearrangements is \[3!\] .
Now, the total number of words that can be formed \[{{=}^{5}}{{C}_{3}}\times 3!=60\] .
In case [ii], we have to find the number of three-letter words that can be formed if repetition is allowed. As repetition is allowed we can take that letter again if it is taken previously. In the first place, we can take a, b, c, d, and e. So, we have 5 intakes in the first place. Suppose, we take “a” in the first place of 3 letter words. We can also take the letter “a” in the second place too. So, in second place we can take a, b, c, d, and e. In the second place, we have 5 intakes. Similarly, in third place, we have 5 intakes.
The total number of words to be formed \[=5\times 5\times 5=125\] .
Note: In this question, the mistake is generally done in finding the number of words that can be formed without repetition. As we have to take three letters out of five given letters, one can directly write \[^{5}{{C}_{3}}\] . But, the three words can rearrange also. So, we also have to think about the rearrangements.
How many 3 letter words can be formed from signature if repetition is not allowed?
=720. Hence, the no. of 3 letter words formed from the word LOGARITHMS without repetition is 720.
How many 3 letter words can be made if letters can be repeated?
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. Show activity on this post.
How many ways 4 letters can be formed from a b/c d/e and f if repetition is allowed?
The answer is 360.
How many 3 letter words can be formed from the letters Abcde if letters can be repeated in a word?
If letters can be repeated as many times as you want, there are 6 options [A, B, C, D, E, or F] for the first letter, second letter, and third letter. Then 63=216 are the number of options for all three-letter-words.