Question

How many 9-letter words can be made from the letters the words ‘COMMITTEE’?
A. $\dfrac{{9!}}{{{{\left( {2!} \right)}^2}}}$.
B. $\dfrac{{9!}}{{{{\left( {2!} \right)}^3}}}$.
C. \[\dfrac{{9!}}{{\left( {2!} \right)}}\].
D. $9!$.

Answer 1

Hint: In this question, we will use the concept of permutations of an arrangement. The number of permutations of n objects, where ${p_1}$objects are of one kind, ${p_2}$ objects are of second kind, ….., ${p_k}$ objects are of ${k^{th}}$ kind and the rest , if any, are of different kind is given by $\dfrac{{n!}}{{{p_1}!{p_2}!.....{p_k}!}}.$

Complete step-by-step answer:
A permutation is an arrangement of a number of objects, any or all taken at a time in a definite order.
We know that the permutations of n objects, where p objects are of same kind and rest are all different is given by $\dfrac{{n!}}{{p!}}.$
Here we have a word ‘COMMITTEE’
In this word there are a total 9 objects (letters) of which there are 2M’s, 2T’s and 2E’s.
i.e. n = 9, ${p_1}$ = 2, ${p_2}$ = 2, ${p_3}$ = 2.
According to the formula of permutations, the number of permutations of 9 objects is,
 $ \Rightarrow \dfrac{{n!}}{{{p_1}!{p_2}!{p_3}!}}.$
Now by putting all the known values, we get
$ \Rightarrow \dfrac{{9!}}{{2!2!2!}}.$
\[ \Rightarrow \dfrac{{9!}}{{{{\left( {2!} \right)}^3}}}.\]
Thus, we can say that \[\dfrac{{9!}}{{{{\left( {2!} \right)}^3}}}\] number of 9-letter words can be made from the letters of the word ‘COMMITTEE’.
Hence the correct answer is option (A).

Note: whenever we ask such types of questions, we will use the method of finding the permutations of n objects. First, we have to find out the total number of objects and then we will find out how many of the objects are of the same kind and how many are different. Then we will use the formula of permutations and through this we will get the answer.