Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 5)
5.
In how many ways can the letters of the word 'LEADER' be arranged?
Answer: Option
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
 Required number of ways = |
6! | = 360. |
| (1!)(2!)(1!)(1!)(1!) |
Video Explanation: https://youtu.be/2_2QukHfkYA
Discussion:
84 comments Page 2 of 9.
Abhishek krishna said:
1 decade ago
Hi Madhu and Shweta, the answer is right if see I tell how the letter E is repeated two times so it has to be divided and rest of letters are only one times occurred, so dividing from 1 did not make any thing.
RDG said:
8 years ago
Sometimes vowel sometimes that way nothing could be understood please explain in sequence. And introduce the differences of combina and permu. Try to teach from a novice perspective.
Nanda said:
9 years ago
Why shouldn't the vowels and consonants be grouped differently ? like LDR (EEA) which makes it 3 + 1 = 4!
And 3!/2! since E has repeated twice like in the previous problems.
And 3!/2! since E has repeated twice like in the previous problems.
SANJAY P said:
9 years ago
I have to plan audit in such a way that 24 activities are to be assessed in 6 different laboratories in a year with 9 assessors. Two assessors are required for on activity.
Parthiban said:
8 years ago
LEADER : totally 6 words so= 6!.
L1, E2, A1, D1, R1 so= 1!*2!*1!*1!*1!.
6!=6*5*4*3*2*1=720.
1!*2!*1!*1!*1!= (1*1) (2*1) (1*1) (1*1) (1*1)=2.
Total words=720/2=360.
L1, E2, A1, D1, R1 so= 1!*2!*1!*1!*1!.
6!=6*5*4*3*2*1=720.
1!*2!*1!*1!*1!= (1*1) (2*1) (1*1) (1*1) (1*1)=2.
Total words=720/2=360.
(1)
Kumar said:
9 years ago
LEADER has 6 words totally.
But letter E repeatedly two times then, to get a required solution we use 6! (total words) /2! (repeated word). So, the answer is 360.
But letter E repeatedly two times then, to get a required solution we use 6! (total words) /2! (repeated word). So, the answer is 360.
Swapnil said:
1 decade ago
In this letter e repeated twice.
So that in permutation swapping of e letter.
Does not form new word and so they are counted.
Once at every case so divided by 2!.
So that in permutation swapping of e letter.
Does not form new word and so they are counted.
Once at every case so divided by 2!.
Manish said:
8 years ago
If vowels point is given then vowel to be treated as one-word & if that one word contains repeat letters they must be divided.
For eg- eeo.
3!/2!.
For eg- eeo.
3!/2!.
Shivaraj kavalaga said:
9 years ago
@Vikram.
EYE
It contains 3 letters i.e 3!
Now take reapeted letters i.e E(2times repeated)=2!,
EYE = 3!/2!,
= 3 * 2 * 1/2 * 1,
= 3.
EYE
It contains 3 letters i.e 3!
Now take reapeted letters i.e E(2times repeated)=2!,
EYE = 3!/2!,
= 3 * 2 * 1/2 * 1,
= 3.
Mukund said:
1 decade ago
Whether we need to divide by 2 or 2!. As in this, it doesn't matter but if "e" was repeated thrice, then do we need to divide numerator by 3 or 3!
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