Question

How many words can be formed from the letters of the word $TRIANGLE?$ In how many of these does the word start with $T$ and end with $E$ ?

Answer 1

Using the concept of counting and permutation we can solve this.
First we solve the first part of the question by permutation and the second part by counting, so that we can understand the concept thoroughly.
Counting technique is the number of ways to choose k objects from a group of n objects.
General form of choose k objects from a group of n objects is $$k!$$$$$$
Permutation is a collection or a combination of objects from a set where the order or the arrangement of the chosen object does matter.

Formula used: General formula for permutation chosen $r$ things from $n$ objects ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$

Complete step-by-step answer:
It is given that the word TRIANGLE contains 8 letters.
Total numbers of ways of permutations are to put these $8$ letters in $8$ places = ${}^8{P_8}$
${\text{TRIANGLE = 8 letters}}$
Here the formula, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
$n =$Total number of letters in the set
$r =$The number of choosing letters from the set
$^8{P_8} = \dfrac{{8!}}{{\left( {8 - 8} \right)!}}$
${}^8{P_8} = \dfrac{{8!}}{{0!}}$
${}^8{P_8} = 8!$
$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
=$40,320$ ways.
Total numbers of ways of permutations are to put these $8$ letters in $8$ places = $40,320$
Now we have to find,
Word beginning with $T$ and ending with $E$ implies that $2$ positions out of $8$ are fixed.
So we need to arrange $6$ letters in $6$ positions.
$T - - - - - - E$
$RIANGL = 6 letters$
$T \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times E$ ways
$T \times 6! \times E$ ways
$6! = 720$ ways
The word start with $T$ and ending with $E$ counted has $720$ways

Note: There is another way of solving this problem, the first part can be done either counting or using permutation. And the second part will be done by either counting or permutation.
Now solving first part by counting
$TRIANGLE = 8 letters$
$- - - - - - - - -$ ways
$8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ ways
$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320$ ways
Second part is solved by permutation.
Total numbers of ways of counting are to put these $8$ letters in $8$ places = $40,320$
The word beginning with $T$ and ending with $E$ implies that $2$ positions out of $8$ are fixed.
$RIANGL = 6letters$
${}^6{P_6} = \dfrac{{6!}}{{(6 - 6)!}}$
${}^6{P_6} = \dfrac{{6!}}{{0!}}$
${}^6{P_6} = 6!$
$6! = 720$ ways
Total numbers of ways of permutations are to put these $6$ letters in $6$ places = $720$