Question

How many different arrangement can be made by using all the letters in the MATHEMATICS?
How many of them begin with $C$? How many of them begin with $T$?

Answer 1

There are 11 words $2M^s,2A^s,2T^s,H,E,I,C,S$.
(i) Hence the number of words by taking all at a time.
$=\frac{11!}{2!2!2!}=\frac{11\times 10\times 9\times 8\times 7\times 6!}{2\times 2\times 2}$
$=990\times 7\times 720=990\times 5040=4989600$.
(ii) To begin with $C$.
Having fixed $C$ at first place we have 100 letters in which 2 are $M^s,2$ are $A^s$ and $2$ are $T^s$ and rest 4 different.
Hence the number of words will be
$\frac{10!}{2!2!2!}=\frac{10\times 9\times 8\times 7\times 6!}{2\times 2\times 2}$
$=90\times 7\times 720=630\times 720=453600$.
(iii) To begin with $T$.
having fixed $T$ in the first place we will have only 10 letters our out of which 2 are $M^s$ and $2$ are $A^s$ and rest six are $H,E,I,C,S$ and $T$.
hence the number of words is
$\frac{10!}{2!2!}=907200$ (i.e. double part of (ii))