Question

In how many different ways can you arrange the letters of the word TRIGONOMETRY?
(a) 59875200
(b) 1232340
(c) 87045670
(d) None of these

Answer 1

Hint: In this given problem, we are given with arranging the letters of the word TRIGONOMETRY. Then we will check how many letters are repeating and how many letters are appearing once. Then, how many letters are there in the word should also be checked. Then, using the formulas, we can get the needed solution.

Complete step by step solution:
According to the question, we are to find out how many different ways you can arrange the letters of the word “TRIGONOMETRY”.
Now, to start with, the answer is the same as that of the other contributions, but I would like to explain a little more how the calculations are obtained.
The number of ways an n-element set can be ordered is:
${{P}_{n}}=n!$
as long as the elements are different from each other.
But suppose the first element is repeated a times. Then, since all the repeats of the first element are indistinguishable from each other, in reality they will only have:
$\dfrac{n!}{a!}$
possible ordinations.
Finally, if the rest of the elements of the set can also be repeated a number of times each, i.e. we have b second elements, c third, etc., then we have:
${{P}_{n\left( a,b,c,... \right)}}=\dfrac{n!}{a!.b!.c!....}$
possibilities of ordering said set of elements.
Now, again, we can also see, T,O and R letters are repeating twice and I,G,N,M,E,Y are taking place only once.
And also, there are 12 letters in the given word.
Thus, we get the number of possibilities $=\dfrac{12!}{\left( 3\times 2! \right)\left( 1!\,\times 5 \right)}=59,875,200$

So, the correct answer is “Option a”.

Note: In this solution, we are only considering with the given letters to us. But there also might be a case when we have to consider the vowels and consonants differently and count the number of words we can make out of it. That should be considered as another typical case of the solution.