Question

Find the number of 4 letter words that can be formed using the letters of the word MATRIX such that:
(i) which contain the letter X?
(ii) and which do not contain the letter X?

Answer 1

Hint: We have to use the letters of the word MATRIX and construct a 4 letter word in such a way that in first case it will contain X and in second case it won’t contain X. To form any word using letters we have to arrange the letters so we have to do the permutations of the letters to construct a word. As you can see that MATRIX contains 6 letters and we are solving the first case so we have to include X. Now, X will be there so we don’t have to select from the MATRIX so we need 3 letters and we have to select 3 letters from the 5 letters and then arrange the 4 letters which contain the selected 3 letters and X. Arrangement of 4 letters is equal to 4!. To solve the second part as we should not include X so now we have to select the 4 letters from 5 letters because we cannot include X and then arrange the 4 letters as 4!.

Complete step-by-step answer:
We have given the word MATRIX and we have to write the number of 4 letter words constructed from these 6 letters of the word MATRIX according to the following conditions:
(i) which contain the letter X?
(ii) and which do not contain the letter X?
We are solving the part (i):
As the 4 letter word contains X it means we don’t have to select X from the 6 letters. Now, X is already selected so from 6 letters one letter (X) gone so we are left with 5 letters and we have to select 3 letters from 5 letters which we are selecting as follows:
${}^5{C}_3$
Now, we have selected the letters but we have to arrange the 4 letters also so the arrangement of 4 letters is done as:
4!
Multiplying the selection and arrangement of 4 letters in a row we get,
${}^5{C}_3(4!)$
We know that:
${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$
Using the above relation in the expansion of ${}^5{C}_3$ we get,
$\begin{array}{rl}& {\displaystyle \frac{5!}{3!2!}}(4!)\\ & ={\displaystyle \frac{5.4.3!}{3!2.1}}\left(4.3.2.1\right)\end{array}$
In the above expression, 3! And 2.1 in the numerator and denominator and we get,
$\begin{array}{rl}& {\displaystyle \frac{5.4}{1}}\left(4.3\right)\\ & =240\end{array}$
Hence, the result of the first part is 240.
We are solving the part (ii):
In this part, we have to find the number of 4 letter words which do not contain the letter X so we have to ignore the letter X in the 6 letters so we have to select 4 letters from 5 letters because we have taken out the letter X from the 5 letters. Now, selecting 4 letters from the 5 letters we get,
${}^5{C}_4$
Arrangement of 4 letter word is equal to:
4!
Multiplying the selection of 4 letters and arrangement of 4 letters we get the number of 4 letter words which do not contain the letter X.
$\begin{array}{rl}& {}^5{C}_4(4!)\\ & ={\displaystyle \frac{5!}{4!1!}}(4!)\end{array}$
In the above expression, 4! Will be cancelled out from the numerator and denominator and we get,
$\begin{array}{rl}& {\displaystyle \frac{5!}{1!}}\\ & =5.4.3.2.1\\ & =120\end{array}$
Hence, we got the result of the second part as 120.

Note: In the above solution, you might have a question that in the first part why we are arranging the 4 letters because we do not select the letter X then why we are including X in arrangement. The answer is that in the selection process we are not selecting X because we have already picked up the letter X from the bunch of 6 letters and kept it aside. And then select the remaining 3 letters from the remaining 5 letters but we have selected 4 letters from the 6 letters whereas in the above problem, there is no specific position is given of X so X can be at any place in the 4 letters arrangement just like other letters.