Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
725 views
in Mathematics by (114k points)
closed by

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


1. \(\dfrac{9!}{2!\times 2! \times 2!}\)
2. 9!
3. \(\dfrac{9!}{2!\times 2!}\)
4. None of these

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 3 : \(\dfrac{9!}{2!\times 2!}\)

Concept:

Suppose a set of n objects has n1 of one kind of object, n2 of a second kind, n3 of a third kind, and so on with n = n1 + n2 + n3  +...+ nk then the number of distinguishable permutations of the n objects is  = \(\rm \frac{n!}{n_1!n_2! ...n_k!}\)

 

Calculation:

The given word is MATHEMATICS

The total number of letters is 11.

Given that we have to form those words which start and end with T.

T _ _ _ _ _ _ _ _ _ _ T

 

Since there are two identical letters and have to be placed in two positions so we can place it only 1 way.

Now we have the remaining 9 positions and the remaining 9 letters are M, A, H, E, M, A, I, C, S

Now in the given word, there are 2T’s, 2M’s and 2A’s, from where 2T’s are already used, so since there are 2M’s and 2A’s

Therefore, the total number of words = \(\dfrac{9!}{2!\times 2!}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...