JEE Main 2018 — Permutation Combination Question with Solution

If all four letters are different then the number of words ${}^5{\mathrm{C}}_4\times 4!=120$ If two letters are $\mathrm{R}$ and other two different letters are chosen from $\mathrm{B},\mathrm{A},\mathrm{C},\mathrm{K}$ then the number of words $={}^4{\mathrm{C}}_2\times \frac{4!}{2!}=72$ If two letters are $\mathrm{A}$ and other two different letters are chosen from $\mathrm{B},\mathrm{R},\mathrm{C},\mathrm{K}$ then the number of words $={}^4{\mathrm{C}}_2\times \frac{4!}{2!}=72$ If word is formed using two $R$ 's and two $A$ 's then the number of words $=\frac{4!}{2!2!}=6$ Therefore, the number of four-letter words that can be formed $=120+72+72+6=270$