Question

# Consider the letters of the word MATHEMATICS. Then the possible number of words

A
without any restriction is 11!212!2!
B
when all the repeating letters are at odd position is 5!×6!2!2!2!
C
when the word starts with consonant is 7×10!8
D
having 7 letters where all letters are unique is 8P7

Solution

## The correct options are A without any restriction is 11!212!2! B when all the repeating letters are at odd position is 5!×6!2!2!2! C when the word starts with consonant is 7×10!8 D having 7 letters where all letters are unique is 8P7(i)Since there are 11 letters in given word which contains 3 pairs of repeated letters,each letter being repeated 2 times, ∴ total number of words without any restrictions =11!2!2!2! (ii)Since there are 6 odd places and 3 pairs of repeated letters (M,A,T) in the given word, therefore all these repeated letters can be arranged at these odd places in 6!2!2!2!, while the remaining letters (H,E,I,C,S)will be arranged in remaining 5 even places in 5! ways. Therefore, required number of ways in which all repeated letters are at odd places is 5!×6!2!2!2! (iii)Words starting with consonants implies that the words can only start from M,T,H,C,S only. Number of words starting with M,A,T=10!2!2! and words staring with H,E,I,C,S=10!2!2!2! Hence, total number of words starting with consonants is : 2(10!2!2!)+3(10!2!2!2!) =10!2!2!(2+32)=7×10!8 (iv)Since we have 8 unique letters, 7 letter words can be formed in 8P7 waysMathematics

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