Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 13)
13.
In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
10080
4989600
120960
None of these
Answer: Option
Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8! = 10080.
(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = 4! = 12.
2!

Required number of words = (10080 x 12) = 120960.

Discussion:
36 comments Page 3 of 4.

Rohit Soni said:   10 years ago
Sir, how have you solved 8!/(2!) (2!) and 4!/2!?

Raji said:   1 decade ago
Why don't take 7c4 instead of 7p4?

VAIBHAV said:   1 decade ago
(AEAI) can also have different arrangement like AAEI, AIAE...etc.

Salman khan said:   1 decade ago
Can we use permutation(nPr) method?

Deepa said:   1 decade ago
Number of ways of arranging these letters = 4!/2! = 2!

And than the answer was = 10080*2*1 = 20160.

Ram Naik said:   1 decade ago
@Wolanyo.

To find out the number of ways to arrange the so that two Ms do not come together.!

Two Ms don't come to gether = Total number of ways - two Ms come together.

Total number ways of arranging MATHEMATICS letters = 11!/(2!*2!*2!) = 4989600.

Two Ms come together = 10!/(2!*2!) = 907200.

Two Ms don't come together = 4989600-907200 = 4082400.

Wolanyo said:   1 decade ago
How many ways can we arrange so that the two Ms do not come together?

Krithika said:   1 decade ago
In the same sum if we are asked to arrange the letters such that the vowels always occur in the same order(i.e- there maybe any no of letters between the vowels, but the order in which the vowels occur in the word should be same).

Usama zaka said:   1 decade ago
@Titesh n Tushar

8! meanx (8*7*6*5*4*3*2*1) which gives us 40320 similarly (2!)(2!) gives us 4. So dividing 40320/4 we get 1008.

Hope you understand.

Riddhi said:   1 decade ago
can't we take it as 7p7*4p4.....


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