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 1 of 4.
Prashanth D said:
6 years ago
GIVEN WORD = M A T H E M A T I C S = TOTAL = 11.
Solution:-
VOWEL FROM THE WORD = A A E I -------------------------> (1)
REMAINING LETTERS IN THE WORD = M T H M T C S ------->(2)
CONSIDER [ A A E I ] M T H M T C S = 1 + 7 = 8! divided by 2! 2! [becus we have M M and T T].
FROM ------->A A E I we get 4! and divided 2! [becus we have two A A].
8! 4! DIVIDED BY 2! 2! 2! =======> 120960.
Solution:-
VOWEL FROM THE WORD = A A E I -------------------------> (1)
REMAINING LETTERS IN THE WORD = M T H M T C S ------->(2)
CONSIDER [ A A E I ] M T H M T C S = 1 + 7 = 8! divided by 2! 2! [becus we have M M and T T].
FROM ------->A A E I we get 4! and divided 2! [becus we have two A A].
8! 4! DIVIDED BY 2! 2! 2! =======> 120960.
(4)
Popra Tetseo said:
6 years ago
Anyone can please solve this for me? In how many ways PENCIL be arranged so that P and C are next to each other?
(2)
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!
Thus, we have MTHMTCS (AEAI).
7! * 4! => 5040 * 24 => 120960.
Make it simple!
(1)
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 isá¡.
10080*12=120960 ways.
# 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 isá¡.
10080*12=120960 ways.
(1)
Shree said:
7 years ago
Thanks all for the clear explanations.
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.
Srav's said:
8 years ago
I am not getting this answer please sir give me a clear explanation.
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!.
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.
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.
Himanshu said:
7 years ago
If vowels never come together then?
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