The number of four letter words that can be formed using the letters of the word BARRACK is:

We look at the seven letters of the word BARRACK. Writing them with their multiplicities, we have $$\text{B, A, R, R, A, C, K}$$ so that $$A$$ appears twice, $$R$$ appears twice, and $$B, C, K$$ appear once each. Thus the multiset of available letters is $$\{A,A,R,R,B,C,K\}.$$ We must count every possible four-letter arrangement (a “word”) that can be formed from this multiset, taking care not to use any letter more often than it exists. The easiest way is to split the counting into mutually exclusive cases based on the repetition pattern inside each word.

Case 1: all four letters are distinct. First we select which four distinct letters will be used. There are five distinct symbols altogether—$$A,B,R,C,K$$—so

$$\binom{5}{4}=5$$

ways to pick the symbols. Once chosen, four different symbols can be arranged in

$$4! = 24$$

orders. Hence this case contributes

$$5 \times 24 = 120$$

words.

Case 2: exactly one letter is repeated once, giving the pattern $$2,1,1.$$ We proceed in two sub-steps.

• Step (a): choose which letter repeats. Only the letters with at least two copies, namely $$A$$ and $$R,$$ qualify, so there are $$2$$ choices.

• Step (b): after fixing the repeated letter, we must pick two other distinct letters different from the repeated one. Excluding the chosen repeater leaves four distinct symbols, and we need any two of them, so

$$\binom{4}{2}=6$$

choices.

Multiplying, the number of different multisets of the form $$\{x,x,y,z\}$$ is

$$2 \times 6 = 12.$$ For any particular multiset containing two identical symbols and two different singles, the total permutations are

$$\frac{4!}{2!} = \frac{24}{2}=12,$$ because dividing by $$2!$$ corrects for the duplication of the repeated letter.

Therefore Case 2 yields

$$12 \times 12 = 144$$

words.

Case 3: two different letters are each repeated twice, giving the pattern $$2,2.$$ Because only $$A$$ and $$R$$ exist in duplicate, the only possible multiset is $$\{A,A,R,R\}.$$

The number of distinct arrangements is

$$\frac{4!}{2!\,2!}= \frac{24}{4}=6.$$

Case 4: any pattern using three or four identical letters is impossible because no letter appears three or more times in BARRACK.

Adding the contributions from every feasible case, we obtain the grand total

$$120 + 144 + 6 = 270.$$

Hence, the correct answer is Option D.