<p>The number of four letter words that can be formed using the letters of the word BARRACK is</p>
Step-by-Step Solution
Key Concept: Count cases based on how many R's and A's are used (since BARRACK has repeated letters: 2 R's, 2 A's, and single B, C, K), then apply permutation formula n!/n₁!n₂!... for each case to account for repetitions within the selected letters.
<p><strong>Step 1:</strong> Identify available letters: B(1), A(2), R(2), C(1), K(1). We need 4-letter words.</p><p><strong>Step 2:</strong> Case analysis based on number of R's and A's:</p><p><strong>Case 1 (2R, 2A, 0 others):</strong> 4!/(2!·2!) = 6</p><p><strong>Case 2 (2R, 1A, 1 other):</strong> Choose 1 from {B,C,K}: 3 ways. Each: 4!/2! = 12. Total: 3×12 = 36</p><p><strong>Case 3 (2R, 0A, 2 others):</strong> Choose 2 from {B,C,K}: C(3,2)=3 ways. Each: 4!/2! = 12. Total: 3×12 = 36</p><p><strong>Case 4 (1R, 2A, 1 other):</strong> Choose 1 from {B,C,K}: 3 ways. Each: 4!/2! = 12. Total: 3×12 = 36</p><p><strong>Case 5 (1R, 1A, 2 others):</strong> Choose 2 from {B,C,K}: C(3,2)=3 ways. Each: 4! = 24. Total: 3×24 = 72</p><p><strong>Case 6 (1R, 0A, 3 others):</strong> Must choose all {B,C,K}: 1 way. 4! = 24. Total: 24</p><p><strong>Case 7 (0R, 2A, 2 others):</strong> Choose 2 from {B,C,K}: C(3,2)=3 ways. Each: 4!/2! = 12. Total: 3×12 = 36</p><p><strong>Case 8 (0R, 1A, 3 others):</strong> Must choose all {B,C,K}: 1 way. 4! = 24. Total: 24</p><p><strong>Step 3:</strong> Total = 6 + 36 + 36 + 36 + 72 + 24 + 36 + 24 = 270</p><p>∴ Answer: C</p>
Correct Answer: C