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.
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)
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.
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)
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.
Ram Naik said:
1 decade ago
@Wolanyo.
To find out the number of ways to arrange the so that two Ms do not come together.!
Two Ms don't come to gether = Total number of ways - two Ms come together.
Total number ways of arranging MATHEMATICS letters = 11!/(2!*2!*2!) = 4989600.
Two Ms come together = 10!/(2!*2!) = 907200.
Two Ms don't come together = 4989600-907200 = 4082400.
To find out the number of ways to arrange the so that two Ms do not come together.!
Two Ms don't come to gether = Total number of ways - two Ms come together.
Total number ways of arranging MATHEMATICS letters = 11!/(2!*2!*2!) = 4989600.
Two Ms come together = 10!/(2!*2!) = 907200.
Two Ms don't come together = 4989600-907200 = 4082400.
Maswoad said:
6 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.
Chandan said:
7 years ago
In the above solution as suggested in the answer section, we are just taking the 4 vowels set at last, but its not said in the question that the vowels set to come together and at last of the word. The words can also be formed as M (AEAI) THMTCS or more which is not considered.
Sabya said:
2 weeks 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).
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.
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).
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