Analyze the letters in BARRACK

The word BARRACK has 7 letters: $B, A, R, R, A, C, K$

Their repetitions are:

• $A$ occurs $2$ times
• $R$ occurs $2$ times
• $B, C, K$ occur $1$ time each

So the available letters are: $\{A,A,R,R,B,C,K\}$

We need the number of 4-letter words

Since letters repeat, we count arrangements case by case according to repetition pattern.

Case 1: All 4 letters distinct

We choose 4 distinct letters from the 5 distinct types $\{A,R,B,C,K\}$: $\binom{5}{4} = 5$

For each choice, number of arrangements: $4! = 24$

Hence total: $5 \times 24 = 120$

Case 2: One letter repeated twice, other two distinct

The only letters that can repeat are $A$ and $R$.

(i) Repeated letter is $A$

Choose 2 other distinct letters from $\{R,B,C,K\}$: $\binom{4}{2} = 6$

Arrange the multiset $(A,A,x,y)$: $\frac{4!}{2!} = 12$

Total: $6 \times 12 = 72$

(ii) Repeated letter is $R$

Choose 2 other distinct letters from $\{A,B,C,K\}$: $\binom{4}{2} = 6$

Arrange $(R,R,x,y)$: $\frac{4!}{2!} = 12$

Total: $6 \times 12 = 72$

So Case 2 total: $72 + 72 = 144$

Case 3: Two letters repeated twice each

This is possible only with $A$ and $R$, giving the multiset: $A,A,R,R$

Number of arrangements: $\frac{4!}{2!2!} = 6$

Total number of 4-letter words

Adding all cases: $120 + 144 + 6 = 270$

Check options

• A: $120$
• B: $144$
• C: $264$
• D: $270$

Therefore, the correct option is: $\boxed{D}$