MHT CET · Maths · Permutation Combination
The number of four letter words that can be formed using letters of the word BARRACK is
- A 120
- B 264
- C 270
- D 144
Answer & Solution
Correct Answer
(C) 270
Step-by-step Solution
Detailed explanation
Word BARRACK has 7 letters in which 'A' and ' \(R\) ' repeats twice.
Case I :
All four letters are different. (B, A, R, C, K)
\(\therefore\) No. of letters \(={ }^5 \mathrm{C}_4 \times 4!=120\)
Case II :
'\(R\)' repeats twice and remaining letter three letters are different (B, A, C, K)
\(\therefore\) No. of letters \(={ }^4 \mathrm{C}_2 \times \frac{4!}{2!}=72\)
Case III :
'A' repeats twice and remaining letter three letters are different (B, R, C, K)
\(\therefore\) No. of letters \(={ }^4 \mathrm{C}_2 \times \frac{4!}{2!}=72\)
Case IV :
Both 'A' and 'R' repeat twice.
\(\therefore\) No. of letters \(=\frac{4!}{2!2!}=6\)
\(\therefore\) Total no. of letters form \(=120+72+72+6\)
\(=270\)
Case I :
All four letters are different. (B, A, R, C, K)
\(\therefore\) No. of letters \(={ }^5 \mathrm{C}_4 \times 4!=120\)
Case II :
'\(R\)' repeats twice and remaining letter three letters are different (B, A, C, K)
\(\therefore\) No. of letters \(={ }^4 \mathrm{C}_2 \times \frac{4!}{2!}=72\)
Case III :
'A' repeats twice and remaining letter three letters are different (B, R, C, K)
\(\therefore\) No. of letters \(={ }^4 \mathrm{C}_2 \times \frac{4!}{2!}=72\)
Case IV :
Both 'A' and 'R' repeat twice.
\(\therefore\) No. of letters \(=\frac{4!}{2!2!}=6\)
\(\therefore\) Total no. of letters form \(=120+72+72+6\)
\(=270\)
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