MHT CET · Maths · Permutation Combination
The number of words that can be formed by using the letters of the word CALCULATE such that each word starts and ends with a consonant, are
- A \(5 \times 7 !\)
- B \(\frac{9 !}{8}\)
- C \(\frac{5 \times 7 !}{2}\)
- D \(20 \times 7 !\)
Answer & Solution
Correct Answer
(C) \(\frac{5 \times 7 !}{2}\)
Step-by-step Solution
Detailed explanation
Word CALCULATE has 9 letters.
Out of which 'C' repeats 2 times,
'A' repeats 2 times,
'L' repeats 2 times,
'E', 'U' and 'T' repeats once.
\(\therefore\) There are 5 consonants and 4 vowels.
Two consonants out of 5 can take start and end position of the word in \({ }^5 \mathrm{P}_2\) ways.
And remaining 7 letters can take remaining 7 positions in 7 ! ways.
Also, 'C', 'A' and 'L' repeats twice each.
\(\therefore\) The required number of words that can be formed \(=\frac{5 p_2 \times 7 !}{2 ! \times 2 ! \times 2 !}=\frac{5 \times 4 \times 3 ! \times 7 !}{3 ! \times 2 \times 2 \times 2}=\frac{5 \times 7 !}{2}\)
Out of which 'C' repeats 2 times,
'A' repeats 2 times,
'L' repeats 2 times,
'E', 'U' and 'T' repeats once.
\(\therefore\) There are 5 consonants and 4 vowels.
Two consonants out of 5 can take start and end position of the word in \({ }^5 \mathrm{P}_2\) ways.
And remaining 7 letters can take remaining 7 positions in 7 ! ways.
Also, 'C', 'A' and 'L' repeats twice each.
\(\therefore\) The required number of words that can be formed \(=\frac{5 p_2 \times 7 !}{2 ! \times 2 ! \times 2 !}=\frac{5 \times 4 \times 3 ! \times 7 !}{3 ! \times 2 \times 2 \times 2}=\frac{5 \times 7 !}{2}\)
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