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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
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The correct answer is:
$\frac{5 \times 7 !}{2}$
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.
$\therefore \quad$ 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.
$\therefore \quad$ 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.
$\therefore \quad$ 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.
$\therefore \quad$ 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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