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 4 of 4.
Deepa said:
1 decade ago
Number of ways of arranging these letters = 4!/2! = 2!
And than the answer was = 10080*2*1 = 20160.
And than the answer was = 10080*2*1 = 20160.
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.
Wolanyo said:
1 decade ago
How many ways can we arrange so that the two Ms do not come together?
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).
Usama zaka said:
1 decade ago
@Titesh n Tushar
8! meanx (8*7*6*5*4*3*2*1) which gives us 40320 similarly (2!)(2!) gives us 4. So dividing 40320/4 we get 1008.
Hope you understand.
8! meanx (8*7*6*5*4*3*2*1) which gives us 40320 similarly (2!)(2!) gives us 4. So dividing 40320/4 we get 1008.
Hope you understand.
Riddhi said:
1 decade ago
can't we take it as 7p7*4p4.....
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