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.
![]() |
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.
(5)
Sabya said:
7 months ago
Yeah, it is correct but there is a simpler way to do this;
Out of 11 places, the last 4 take it as vowels so there would be 4! places and the remaining 7 places will come together as 7! ways.
Hence, 7!*4!=120960 (that's the answer).
Out of 11 places, the last 4 take it as vowels so there would be 4! places and the remaining 7 places will come together as 7! ways.
Hence, 7!*4!=120960 (that's the answer).
(3)
Popra Tetseo said:
7 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)
Srav's said:
8 years ago
I am not getting this answer please sir give me a clear explanation.
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.
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.
Sourabh sharma said:
6 years ago
Sir how to find the total arrangements of word ;mathematics, of both A and both M together
AA. MM
AA. MM
Ritesh kumar said:
1 decade ago
Sir I want to know 10080 and 8/(2!) (2!) how comes.
Maswoad said:
7 years ago
@Popra.
In the arrangement of letters of word PENCIL, P and C next to each other means P and C always come together and only one possible arrangement of P and C letter that is PC, CP is wrong. Now think of available letters E, N, I, L, [PC].
So possible arrangement = 5! => 120.
In the arrangement of letters of word PENCIL, P and C next to each other means P and C always come together and only one possible arrangement of P and C letter that is PC, CP is wrong. Now think of available letters E, N, I, L, [PC].
So possible arrangement = 5! => 120.
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