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

The correct option is A

$720$

Step 1: Use combination formula

In the word $LOGARITHMS$ there are $10$ unique letters which are, $A,G,H,I,L,M,O,R,S$and $T$.

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

$\begin{array}{l}{}^n{C}_r=\frac{n! }{r! (n-r)! }\end{array}$

where n! is

$n!=n\times (n–1)\times (n–2)\times (n–3)\times \dots \dots ..\times 3\times 2\times 1$

So, three letters out of $10$ unique letters can be selected in ${}^{10}C_3$ ways.

By using the above formula we get

$\begin{array}{l}{}^{10}{C}_3=\frac{10!}{3!(10-3)!}\end{array}$

$\begin{array}{l}=\frac{10!}{3!\left(7\right)!}\end{array}$

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 $={}^{10}C_3\times 3!$

$\therefore \begin{array}{l}{}^{10}{C}_3\times 3!=\frac{10!}{3!\left(7\right)!}\times 3!\end{array}$

$\begin{array}{l}=\frac{10!}{7!}\end{array}$

$\begin{array}{l}=\frac{10\times 9\times 8\times 7!}{7!}\end{array}$

$\begin{array}{l}=10\times 9\times 8\end{array}$

$=720$

Hence, the word $LOGARITHMS$ if repetition of letters is not allowed can form $720$ number of $3$-letter words.