Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 2)
2.
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
Answer: Option
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
Video Explanation: https://youtu.be/WCEF3iW3H2c
Discussion:
97 comments Page 2 of 10.
Deepanesh said:
6 years ago
Why we are taking possibilities of vowels too?
(1)
Saravanakumar said:
7 years ago
LEADING.
Total vowels : E , A, I
remaining : LDNG
assume : EAI we can arrange (EAI)L, L(EAI), D(EAI), N(EAI),G(EAI)
so, total remaining is 4 so, 5!
and possibilities of arranging refer "assume" , 3!
then answer is : 5!*3!
120 * 6=720.
Total vowels : E , A, I
remaining : LDNG
assume : EAI we can arrange (EAI)L, L(EAI), D(EAI), N(EAI),G(EAI)
so, total remaining is 4 so, 5!
and possibilities of arranging refer "assume" , 3!
then answer is : 5!*3!
120 * 6=720.
(1)
Sri said:
1 decade ago
As vowels are together take (EAI) as single letter i.e. , total no of letters are 5 (L, N, D, G, {EAI}).
No of ways can arrange these 5 letters are 5! ways.
Now we arranged 5 letters (L, N, D, G, {EAI}).
Next we have to arrange E, A, I (they may be EAI/EIA/AEI/AIE/IAE/IEA).
All these combinations imply that vowels are together.
So we have to multiply 5! and 3!.
No of ways can arrange these 5 letters are 5! ways.
Now we arranged 5 letters (L, N, D, G, {EAI}).
Next we have to arrange E, A, I (they may be EAI/EIA/AEI/AIE/IAE/IEA).
All these combinations imply that vowels are together.
So we have to multiply 5! and 3!.
(1)
Subbu said:
1 decade ago
7!=5040
(1)
Madhusudan said:
1 decade ago
Could you kindly let me know what is ( ! ).Howe 5! = 120 ?
(1)
Sundar said:
1 decade ago
@Madhusudan
5! = 5 Factorial = 1 x 2 x 3 x 4 x 5 = 120
5! = 5 Factorial = 1 x 2 x 3 x 4 x 5 = 120
(1)
Jessie said:
1 decade ago
7 letter word = LEADING
CONDITION = VOWELS TO BE TOGETHER, HENCE (EAI) TO FOR A WORD
SO NO. OF WORDS = L,(EAI),D,N,G = 5
permuation to arrange 5 letters = nPr= n!/(n-r)!=5!/0!=5!
0! is assumed to be 1!)
EAI can be arranged among each other in = nPr = 3!/(3-3)= 3!
hence 5! x 3! = 120 x 6 = 720
CONDITION = VOWELS TO BE TOGETHER, HENCE (EAI) TO FOR A WORD
SO NO. OF WORDS = L,(EAI),D,N,G = 5
permuation to arrange 5 letters = nPr= n!/(n-r)!=5!/0!=5!
0! is assumed to be 1!)
EAI can be arranged among each other in = nPr = 3!/(3-3)= 3!
hence 5! x 3! = 120 x 6 = 720
(1)
Xyz said:
9 years ago
How many letters can be formed from COMBINATION if vowels are kept together?
Please give the solution.
Please give the solution.
Nabin shah said:
8 years ago
Why do 120 and 6 multiply?
Dhana said:
8 years ago
Word : leading
condition : vowels together
eai-3!
ldng&eai-5!
(ldng)&(eai)-2!
Whether the last condition is valid?
Give explanation.
condition : vowels together
eai-3!
ldng&eai-5!
(ldng)&(eai)-2!
Whether the last condition is valid?
Give explanation.
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