Question

The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as end with T are
A) 80720
B) 90720
C) 20860
D) 37528

Answer 1

For this question, use permutation and combination to solve this question while keeping both the T in the words at the front and back of the words you form. Also keep in mind that repeated words will not be counted so you need to think how you will eliminate them.

Formulas used:
When finding out the number of combinations, it can be done by doing factorial.
let all possible combinations of a word with m alphabets be N. Then,
N=m!

Complete step by step solution:
For this question, let us assume that the answer to our question is X.
First course of action will be to dissect the word MATHEMATICS. Writing the alphabets in ascending order, we have
MATHEMATICS = A A C E H I M M S T T
It has 11 alphabets. Out of those 11, 2 are fixed which are both the T, one at the beginning and another at the end, leaving us with 9 alphabets. Thus 9! will be the number of words which can be made.
There is one more trick the question has up its sleeve though. Observe that there are two sets of M and A in the words and so they will form repetitive words. The number of words to be removed because of two M will be 2! and similarly, for two A, it will be 2! which together will be $2! \times 2!$ which we will divide the equation with.
Thus, our desired answer will be
$\ X = \dfrac{{9!}}{{2! \times 2!}} \\\ = \dfrac{{362880}}{{2 \times 2}} \\\ = \dfrac{{362880}}{4} \\\ = 90720 \\\ \$

Note:
For finding the factorial, we multiply the number with all preceding numbers. For example, $5! = 5 \times 4 \times 3 \times 2 \times 1$. Also, we did not consider the repetition of T and that is because its repetition has already been ignored when we fixed both the T at the beginning and end, removing any possibility of them interchanging and then we worked with only 9 alphabets, ignoring both the T as they were already fixed.