Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 10)
10.
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
Answer: Option
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.
Discussion:
59 comments Page 1 of 6.
Gayathri said:
1 decade ago
Why we use permutation concept instead of combination.
Manasa said:
1 decade ago
@Gayathri
In this case we are not selecting any letters. Just we are arranging the letters according to given condition.
In this case we are not selecting any letters. Just we are arranging the letters according to given condition.
Kesav said:
1 decade ago
Can any one solve this problem for me?
1! + 2! + .... + 50!=?
a) 3.1035*10^64
b) 2.1021*10^65
c) 3.1035*10^63
d) 3.1035*10^62
1! + 2! + .... + 50!=?
a) 3.1035*10^64
b) 2.1021*10^65
c) 3.1035*10^63
d) 3.1035*10^62
Riddhi said:
1 decade ago
@kesav:plz give me ans. Is it b?
Vishnu said:
1 decade ago
It is asked in TCS apti test.
Gautam said:
1 decade ago
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3)
=3!/0! ,=3*2*1= 6 {we define 0! = 1}
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3)
=3!/0! ,=3*2*1= 6 {we define 0! = 1}.
Total number of ways = (6 x 6) = 36.
Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3)
=3!/0! ,=3*2*1= 6 {we define 0! = 1}
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3)
=3!/0! ,=3*2*1= 6 {we define 0! = 1}.
Total number of ways = (6 x 6) = 36.
Nikhil said:
1 decade ago
Vowel can even be put at the 7th place, not considered above, which will increase the number of possible arrangements. Please clarify?
Athu said:
1 decade ago
What is odd positions?
Drushya said:
1 decade ago
@Athu.
Odd position means its like odd no.
For example : 1 2 3 4 5 6.
In this case odd position is :1 3 5 & even position : 2 4 6.
Odd position means its like odd no.
For example : 1 2 3 4 5 6.
In this case odd position is :1 3 5 & even position : 2 4 6.
Jinal said:
1 decade ago
Vowels can also take position 7 (the last position) ?
V C V C V C V.
So the answer would be 3!*4C3 = 24.
Is this correct?
V C V C V C V.
So the answer would be 3!*4C3 = 24.
Is this correct?
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