Question

The number of four-letter words that can be formed using the letters of the word BARRACK is: -
(a) 144
(b) 120
(c) 264
(d) 270

Answer 1

Form four cases such that, in the first case, all the letters are different in the second case two different letters and 2 A’s, in the third case two different letters and 2 R’s, and in fourth case 2 A’s and 2 R’s. Find the total number of ways of selecting the letters by using the formula ${}^{n}{{C}_{r}}$ and arranging them by using the formula n!. Add all the ways obtained in the four cases and that will be our answer.

Complete step-by-step solution
We have been provided with the word BARRACK and we have to find the total number of words that can be formed by using the four letters of this word. So, the following four cases arise: -
1. Case (i): - Considering all four letters different.
Here, the different letters from which we can choose 4 letters are: - B, A, R, C, K. So, we have to select 4 letters from these 5 letters and arrange them at four different places. Therefore, we have,
Number of ways to select four letters from five letters = ${}^{5}{{C}_{4}}$.
So, the number of ways to arrange them at 4 different places = 4!
Therefore, the number of arrangements = ${}^{5}{{C}_{4}}\times 4!=5\times 4!$
2. Case (ii): - Considering 2 A’s and two different letters.
So, here we have considered two letters as A and the other two letters must be different. Therefore, we have to choose the other two letters from B, R, C, K. So, we have,
Number of ways to select two letters from four letters = ${}^{4}{{C}_{2}}$.
So, the number of ways to arrange 2 A’s and 2 different letters = $\dfrac{4!}{2!}$. Here, 2! Is in the denominator because A is repeating twice.
Therefore, the number of arrangements = ${}^{4}{{C}_{2}}\times \dfrac{4!}{2!}$.
3. Case (iii): - Considering 2 R’s and two different letters.
So, here we have considered two letters as R and the other two letters must be different. Therefore, we have to choose the other two letters from B, A, C, K. So, we have,
Number of ways to select two letters from four letters = ${}^{4}{{C}_{2}}$.
So, the number of ways to arrange 2 R’s and 2 different letters = $\dfrac{4!}{2!}$. Here, 2! Is in the denominator because R is repeating twice.
Therefore, the number of arrangements = ${}^{4}{{C}_{2}}\times \dfrac{4!}{2!}$.
4. Case (iv): - Considering 2 A’s and 2 R’s.
Here, we have selected four letters in which there are 2 A’s and 2 R’s. So, the number of ways to arrange 2 A’s and 2 R’s.
So, the number of ways to arrange 2 A’s and 2 R’s = $\dfrac{4!}{2!\times 2!}$, because both A and r are repeated twice.
Now, on observing all the cases we can say that the overall arrangements possible will be the sum of the number of arrangements in each case.
$\Rightarrow$ Overall arrangements possible = total number of words formed
$\Rightarrow$ Overall arrangements possible = $\left( 5\times 4! \right)+\left( {}^{4}{{C}_{2}}\times \dfrac{4!}{2!} \right)+\left( {}^{4}{{C}_{2}}\times \dfrac{4!}{2!} \right)+\dfrac{4!}{2!\times 2!}$
$\Rightarrow$ Overall arrangements possible = 270
Hence, option (d) is the correct answer.

Note: One must remember that for selecting ‘r’ things from a total of ‘n’ things we use the formula ${}^{n}{{C}_{r}}$ and for arranging ‘n’ things at ‘n’ places we use the formula n!. If any letter is repeating ‘k’ times then the number of possible arrangements is divided by k! to get the answer. As you can see that case (ii) and case (iii) are similar, the only difference is that we have selected 2 A’s in case (ii) and 2 R’s in case (iii).