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 2 of 4.
Deepa said:
1 decade ago
Why dont we use nPr here...as we were using that in last sum. whats the need otherwise where to use nPr..can anybody explain plz ?
Rakesh said:
7 years ago
But here are 6 odd spaces and 4 vowels shouldn't we consider that? Please explain to me.
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.
Gouthami said:
7 years ago
My answer is 11!/2!2!2!. Is it is wrong, please explain the answer I have a dought.
Shree said:
7 years ago
Thanks all for the clear explanations.
Krishan Senarath said:
8 years ago
If four countries are contesting for 5 cups in a competition, how many results can be there?
Rachael said:
8 years ago
How can we know that how to separate the vowels?
Himanshu said:
8 years ago
If vowels never come together then?
Tushar said:
1 decade ago
I want to know 8/(2!) (2!)=10080 how comes.
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.
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