A) 144
B) 120
C) 264
D) 270
Correct Answer: D
Solution :
Case 1: If all four letters are different then the number of words \[{{=}^{5}}{{C}_{4}}\times 4!=120\] Case 2: If 2 letters are R and other 2 different letters are chosen from B, A, C, K then the number of words\[{{=}^{4}}{{C}_{2}}\times \frac{4!}{4!}=72\] Case 3: If 2 letters are A and other 2 different letters are chosen from B, R, C, K then the number of words\[{{=}^{4}}{{C}_{2}}\times \frac{4!}{4!}=72\] Case 4: when word is formed using \[2{{R}^{'}}s\]and\[2A's=\frac{4!}{2!2!}=6\] Then the number of four-letter words that can be formed \[=120+72+72+6=270\]
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