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 2 of 4.

Rachael said:   7 years ago
How can we know that how to separate the vowels?

Himanshu said:   7 years ago
If vowels never come together then?

Getnet Tadesse said:   8 years ago
MATHEMATICS.

# The consonants MTHMTCS are considered as 7 letters.

# The four vowels AEAI are considered as one letter since they are supposed to be arranged all together.

# The 7 consonant + 1 group of vowels= 8 letters.

# If there are no repeated letters in these 8 letters, the total ways of arrangements would be 8 != 40320.

# But in the 8 letters we set up above we know M and T are happened to occur twice. Since a letter repeated twice has two arrangements M and T add up to have 4 repeated arrangements. These 4 repeated arrangements will divide the total 40320 arrangements in to 4. Hence, 40320÷4= 10080.

# The AEAI letter will form (4!) 24 arrangements if there is no letter repeated. But A is repeated twice which divide the 24 arrangements by 2. Hence, we do have 12. arrangements.

# Finally the total number of ways to arrange is፡.

10080*12=120960 ways.
(1)

ANKUR said:   8 years ago
In the word MATHEMATICS, We treat the vowel as one letter. Thus we have MTHMTCS (AEAI). Now we have t arrange 7 letters, out of which M and T occur twice. So no of arranging these letters = 7!/2!*2! = 1260.

Now, AEAI has 4 letters in which A occurs twice.

So, no of ways of arranging these letters = 4!/2!=12.
Required no of words= 1260*12=15120.
So answer is none of these.

Kerese said:   8 years ago
@ALL.

We should notice that in the word "mathematics'", m and t occurred twice 8!/2!2!.

Srav's said:   8 years ago
I am not getting this answer please sir give me a clear explanation.

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.

Baskar said:   9 years ago
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

7! * 4! => 5040 * 24 => 120960.

Make it simple!
(1)

Sudama said:   9 years ago
Can we use permutation(nPr) method? & also Used Other Methods.

Pruthvi said:   9 years ago
This answer is wrong we have to multiply by 2 at last which indicates arrange of vowel and other group.


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