Question

How many different arrangements can be made, with the letters of the word MATHEMATICS? In how many of these arrangements, vowels occur together?

Answer 1

Answer:

(a) 4989600

(b) 120960

Step-by-step explanation:

Hi,

Given word is MATHEMATICS,

There are 11 letters in the word out of which M is repeating twice ,

A is repeating twice, T is repeating twice and the rest 5 being distinct.

So, total number of ways of arranging letters of the word

MATHEMATICS are 11!/2!2!2! = 4989600

(b) There are 2 A's 1 E 1 I which form vowel group, so there are 4

vowels in the word MATHEMATICS,

Treating all the vowels as 1 group and the rest 7 letters with M

repeating twice and T repeating twice, the number of arrangements

are 8!/2!2! , but the vowels can permute between themselves in

4! /2! = 12 ways(Dividing by 2 , since A's can permute in 2! ways

resulting in same arrangement),

So total number of ways in which vowels occur together

are 8!/4*12

= 3*8!

= 120960

Hope, it helps !

Answer 2

AnswEr:

There are 11 letters in the word 'MATHEMATICS' of which two are M's, two are A's, two are T's and all other are distinct. So,

• $\text{Required\:number\:of\:arrangements:-}$

$\qquad \tt \frac{11}{2 ! \times 2 ! \times 2 !} = 4989600$

There are 4 vowels viz. A,E,A,I. Considering these four vowels as one letters we have 8 letters (M, T, H, M, T, C, S and one letter obtained by combining all vowels ), out of which M occurs twice, T occurs twice and the rest all different. These 8 letters can be arranged in :

$\qquad \tt \frac{8 !}{2 ! \times 2 !}$

ways.

• But the four vowels (A, E, A, I ) can be put together in :

$\qquad \tt \frac{4 !}{2 !}$

ways.

° Hence, the total number of arrangements in which vowels are always together-

$\qquad \tt \frac{8 !}{2 ! \times 2 !} \times \frac{4 !}{2 !} \\ \\ \qquad \tt = 10080 \times 12 = 120960.$