How many 4 letters words with or without meaning can be formed out of the letters of the word LOGARITHMS if repetition of letters is not allowed?

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.

11.	In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Answer: Option A Explanation: Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = 7 x 6 x 3 = 63. 2 x 1

13.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A. 10080 B. 4989600 C. 120960 D. None of these Answer: Option C Explanation: In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. Number of ways of arranging these letters = 8! = 10080. (2!)(2!) Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! = 12. 2! Required number of words = (10080 x 12) = 120960.

Join The Discussion

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) 40B) 400C) 5040D) 2520

Answer

Verified

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.

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