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?
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.
Kerese said:
8 years ago
@ALL.
We should notice that in the word "mathematics'", m and t occurred twice 8!/2!2!.
We should notice that in the word "mathematics'", m and t occurred twice 8!/2!2!.
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.
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.
Pranjali said:
9 years ago
Don't you think the overall answer should be multiplied by 2? MTHMTCS (AEAI) and (AEAI) MTHMTCS. I am little confused here. Anyone help me to get this.
Arti said:
2 decades ago
How we consider 8 letter?
Sudama said:
9 years ago
Can we use permutation(nPr) method? & also Used Other Methods.
Pruthvi said:
10 years ago
This answer is wrong we have to multiply by 2 at last which indicates arrange of vowel and other group.
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?
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