Question

Let "k" be the number of permutations can be made out of the letters of the word $TRIANGLE$.Let "m" be the number of permutations that begin with $T$ and end with $E$. Find $\frac{k}{8m}$.

Answer 1

The word $TRIANGLE$ has eight different letters, which can be arranged themselves in $8!$ ways.
$\therefore$ Total number of permutations $=$ $8!=40320.$
Again, when $T$ is fixed at the first place and $E$ at the last place, the remaining six can be arranged themselves in $6!$ ways.
$\therefore$ The number of permutations which begin with $T$ and end with $E$ $=$ $6!$ $=$ $720$
Now $k=40320$ and $m=720$
$\therefore \frac{k}{8m}=7$