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?
360
480
720
5040
None of these
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 3 of 10.

Raj said:   1 decade ago
Friends, How do you say this question is permutation.

Mahanthesh said:   1 decade ago
Hi guys. As per the question, it s mentioned that all vowels should be together, but it has not been mentioned that it should be EAI. According to me these 3 vowels can be arranged in 3*2=6 ways. And the remaining letters LDNG can be arranged in 4*3*2*1= 24 ways. Can anyone explain me about this please ?

Sandeep said:   1 decade ago
can any one explain ? y 5!*3! & y not like this 5!+3!

Srk said:   1 decade ago
@Mahantesh.

1. There is given a word LEADING in this LDNG (consonents) , EAI (vowels).

2. They asked here vowels always come together and so we should have to take LDNG (EAI).

3. We can take as (4+1) ! i.e. here we have to take vowels as together as 1. So we have a chance of 5!.

4. But with in vowels we have many arrangements i.e 3!

5. Finally 5!*3!

Shafi said:   1 decade ago
@Mahantesh,

You are correct 3! and 4!, because you spitted vowels and consonants,

But to combine them vowels+consonants.

i.e. (4 consonants + 1 set of vowels) = 5!

And (3 vowels {1 set}) = 3!

So 5!*3! = 5x4x3x2x1x3x2x1=720.

Adhityasena said:   1 decade ago
I get the answer to be 2*720. Because, since the vowels must come together in the word LEADING, which actually has 7 letters, E and A can be taken as one unit, so now we have L, EA, D, I, N, G to be arranged and which can be done in 6! ways. But among E and A there are 2 arrangements namely EA and AE, So the final answer is 2*720. Am I right !

Chetas said:   1 decade ago
@Adhityasena.

You have not considered a vowel 'I'. 'EAI' is to be taken as one unit.

Sanjay(9776387850) said:   1 decade ago
Hi Guys, at first I tell you how can you know is this permutation or combination. Permutation means arrange(row/column) and Combination means selection(group).

i.e. Number and Word are perm. Playing 11 and committee are combi. This is a word so this is perm. You should know the formula that m different objects are alike and n different object are alike if we arrange all the m+n objects such that n objects are always together=(m+1)!*n!

Here n = Vowel = 3 and m = Con. = 4.

So = (4+1)!*3! = 5!*3!.

= 120*6 = 720.

Chithra said:   1 decade ago
As we known very well that * is consider to be AND, + is consider to be OR. Consider the 1st example 7 men and 6 women problem we taken as (7c3*6c2) + (7c4*6c2) + (7c5).

We can also said this as (7c3 AND 6c2) OR (7c4 AND 6c2) OR (7c5).

CONDITION -> 5 people need to select. We should take 3 men from 7 men and 2 women from 6 women or other option is to take 4 men from 7 men and 1 women from 6 women.

That's why we are using * at the place of AND, + at the place of OR. Similarly, we need to consider both the consonant AND vowel not consonant OR vowel.

So we use 5!*3!.

Taku mambo bellnuisemarbel said:   1 decade ago
Please help me with this; whenever I hear of probability that a variable is being selected what should I think of?


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