Question

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

• A. without any restriction is $\frac{11!}{212!2!}$
• B. when all the repeating letters are at odd position is $5!\times \frac{6!}{2!2!2!}$
• C. when the word starts with consonant is $\frac{7\times 10!}{8}$
• D. having $7$ letters where all letters are unique is${}^8{P}_7$

Answer 1

Answer: (A), (B), (C), (D)

having $7$ letters where all letters are unique is${}^8{P}_7$

$\left(1\right)$ Since, there are $11$ letters in given word which contains $3$ pairs of repeated letters, each letter being repeated $2$ times.
$\therefore$ Total number of words without any restrictions $=\frac{11!}{2!2!2!}$

$\left(2\right)$ 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 $\frac{6!}{2!2!2!}$, 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 $5!\times \frac{6!}{2!2!2!}.$

$\left(3\right)$ Words starting with consonants implies that the words can only start from $M,T,H,C,S$ only.

Number of words starting with $M,A,T=\frac{10!}{2!2!}$
and words staring with $H,E,I,C,S=\frac{10!}{2!2!2!}$

Hence, total number of words starting with consonants is :
$2\left(\frac{10!}{2!2!}\right)+3\left(\frac{10!}{2!2!2!}\right)$

$=\frac{10!}{2!2!}(2+\frac{3}{2})=\frac{7\times 10!}{8}$

$\left(4\right)$ Since, we have $8$ unique letters, $7$ letter words can be formed in ${}^8{P}_7$ ways.