Aptitude - Permutation and Combination - Discussion

Correct me if I'm wrong but,

There are 7 letters:

L E A D I N G

3 vowels and 5 non-vowels.

I got the part where we need to permutate 3 and 5, resulting with 3!*5!. But aren't there also several other positions that these vowels could be positioned?

For Example:

AIELDNG is one factor, another factor is LAIEDNG, another factor is LDAIENG.

As you can see, the positions of the 3 vowels with each other are the same, and the order of non-vowels in the word is also the same, but I only changed the position of the starting point for the vowel permutations, starting from the first position, to the second, to the third. So, I believe that the answer should be 3!*5!*5. Since there are 5 different ways you could represent the same order of vowels in different positions with the same order of non-vowels.

Please do correct me if I'm wrong or if I misunderstood the question

@Tom and @Shantha.

Whenever there is a reference to some arrangement it is permutation. Whenever there is a reference to some selection it is combination.

The given question requires us to find the number of ways in which the word LEADING can be arranged with the condition that the vowels (EAI) always be together. Thus we need to apply the concept of permutation.

Here, since the vowels EAI must always be together we consider it as a single word (EAI).

Thus, LDNG (EAI) make a 5 letter word.

It can be arranged in 5p5 ways = 5!ways.

Now, since EAI can arrange itself in 3p3 or 3! ways, the word LDNG (EAI).

Can thus be arranged or PERMUTED in 3!*5! ways = 720 ways.

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!.

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.

Hi friends.

We know that formula n!=n (n-1) (n-2).....3.2.1. Suppose there n way to choose first element (since there are n elements).

After that there are n-1 ways to choose second element because already we choose one element from n elements that's why we are assuming this way. Similarly n-2 ways to chose the third element..etc it's going like this.

n!=n (n-1).

n!=n (n-1) (n-2) if n>2 or equals 2.

n!=n (n-1) (n-2) (n-3) if n>3 or equals 3.

Hope you understood.

@Avishek

No. of vowels is 3Es and the can only be together in this pattern {EEE}. There is only 1 way of choosing so 1! = 1 * 1 = 1.

We are to consider 3Es as a single alphabet because they are together. So we have L, M, N, T,{EEE} ie 5 letters now in all.

Now of way of choosing 5 letters = 5! (5 * 4 * 3 * 2 * 1) = 120.
Finally, we multiply 120 * 1 = 120 ie. the number of ways of arranging the letters so that the vowels are always together.

Friends in the word LEADING.
there are three vowels A,E,I
With AEI you can form 3! arrangements ie 6 arrangements
As we want the vowels to come together they may be placed at any of the 5 locations on the word LEADING.

So there are 5! arrangements for this.
5! X 3!=720.
The 3 vowels are considered as a single word to simply this process of counting the number if locations.

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!.

@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!

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 !