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 1 of 4.

Arti said:   1 decade ago
How we consider 8 letter?

Sahithya said:   1 decade ago
Hi Arti,.

After considering vowels as 1 set, remaning consonents and the vowel set => M+T+H+M+T+C+S+(AEAI) => 8 letters {because we hav to consider that the vowels always come together}.

I hope you will understand my answer.

Ritesh kumar said:   1 decade ago
Sir I want to know 10080 and 8/(2!) (2!) how comes.

Deepa said:   1 decade ago
Why dont we use nPr here...as we were using that in last sum. whats the need otherwise where to use nPr..can anybody explain plz ?

Tushar said:   1 decade ago
I want to know 8/(2!) (2!)=10080 how comes.

Mini said:   1 decade ago
@ Tushar
After considering vowels as 1 set, remaning consonents and the vowel set => M+T+H+M+T+C+S+(AEAI) = 8 letters {because we hav to consider that the vowels always come together}.
Now 8!/(2!)(2!)
=8*7*6*5*4*3*2*1*/(2*1)*(2*1)=10080
here the (2!) used becoz in the word M+T+H+M+T+C+S+(AEAI)the M occurs twice so one (2!) is for dat and the other (2!) is for T which also occurs twice, while all the other letters including (AEAI)occur only once.(AEAI) occurs only once bcoz we r cosidering it as one letter.
And for arranging (AEAI)= 4!/2! [bcoz here A occurs twice]
=4*3*2*1/2*1=12
Req. no. of words=10080*12=120960

The formula vich v used in this qestion is
= n!/(P1!)(P2!)....(Pr!) where n is the no. of letters nd P is the no. of occurrence of each letter.

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

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.

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).

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


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