Question

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. 720
b. 420
c. 5040
d. None of these

Answer 1

Hint: We start solving the problem by finding the total unique letters present in LOGARITHMS. We then find the total no. of ways to choose 3 letters from all the unique letters obtained from LOGARITHMS. We then find the total no. of ways to arrange these selected letters. We then multiply the total no. of ways to choose and total no. of ways to arrange the selected letters to get the required value.

Complete step-by-step solution
In word LOGARITHMS there are 10 unique letters (i.e A, G, H, I, L, M, O, R, S, T).
Now we have to form a 3-letter word with or without meaning and it is given that repetition of letters is not allowed which means you can’t use the same letter more than one time to form 3-letter words.
Suppose in this question if repetition is allowed then you can use any letter more one time to form 3-letter words.
Now, we know that number of combinations of r objects chosen from n objects when repetition is not allowed is given by:
$${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$$-(1)
Where n! is defined as
$n!=n\times (n-1)\times (n-2)\times (n-3)\times -\times 3\times 2\times 1$
So,
3 - letters out of 10 unique letters can be selected in $${}^{10}{C}_3$$ ways. By using formula (1), we get
$\Rightarrow {}^{10}{C}_3={\displaystyle \frac{10!}{3!\times (10-3)!}}$.
$\Rightarrow {}^{10}{C}_3={\displaystyle \frac{10!}{3!\times 7!}}$.
In general, n distinct objects can be arranged in n!. Here we selected 3-letter out of 10 unique letters, and these selected letters can be arranged in 3!
Total number of 3 letter word =${}^{10}{C}_3\times 3!$
$$\Rightarrow {}^{10}{C}_3\times 3!={\displaystyle \frac{10!}{3!\times 7!}}\times 3!$$.
$$\Rightarrow {}^{10}{C}_3\times 3!={\displaystyle \frac{10!}{\text{⧸}3!\times 7!}}\times \text{⧸}3!$$.
$$\Rightarrow {}^{10}{C}_3\times 3!={\displaystyle \frac{10!}{7!}}$$.
We know that $$n!=n\times (n-1)\times (n-2)\times (n-3)!$$ -(2)
So, using equation (2) and canceling 7! from numerator and denominator we get,
$\Rightarrow {\displaystyle \frac{10!}{7!}}={\displaystyle \frac{10\times 9\times 8\times 7!}{7!}}$.
$\Rightarrow {\displaystyle \frac{10!}{7!}}=10\times 9\times 8$.
$\Rightarrow {\displaystyle \frac{10!}{7!}}=720$.
Hence, the no. of 3 letter words formed from the word LOGARITHMS without repetition is 720.
Hence the correct option of this question is option (a).

Note: We should not stop solving the problem after finding the total no. of ways to choose 3 letters, as arranging these letters differently will give us different words. We can also solve this problem by taking 3 empty boxes and check what will be the total favorable cases. Similarly, we can also expect problems to find the total no. of words with repetition allowed.