Re-arranging letters

[papgust]

In how many ways can the letters of the word "DOUBLE" be re-arranged such that the order in which the vowels appear in the word does not change?

Solution given is 6!/3!. Why are we dividing by 3! here? Please help me understand

[poojaranga]

My thought would be-
O, U & E need to be in the same place, (as indicated in the question). Hence, the total number of possibilities would need to be restricted to the other 3 letters- which would be 3!

[uttam.albela]

In how many ways can the letters of the word "DOUBLE" be re-arranged such that the order in which the vowels appear in the word does not change?

Solution given is 6!/3!. Why are we dividing by 3! here? Please help me understand

We have 6 letters in DOUBLE, so there are 6 places to be filled by these 6 letters..

Let us first think of arranging DBL.

Out of 6 places, select 3 places. No of ways = 6C3
No. of ways to Arrange these letters DBL in selected 3 places = 3!
Now we are left with 3 places where we need to put OUE in this same order.
No of ways to do this = 1

So total number of ways = 6C3 * 3! * 1 = 6!/3! = 120

Let me know if any part of it is not clear to you.

[papgust]

Out of 6 places, select 3 places. No of ways = 6C3
No. of ways to Arrange these letters DBL in selected 3 places = 3!
Now we are left with 3 places where we need to put OUE in this same order.
No of ways to do this = 1

So total number of ways = 6C3 * 3! * 1 = 6!/3! = 120

Let me know if any part of it is not clear to you.

Thanks for the post! I'm still not clear why we are doing 6C3 here. Otherwise i'm clear abt rest of the steps.

My method:
Consider vowels first and this can be arranged in only 1 way.
Then remaining three places can be arranged with 3 remaining letters in 3! ways.
Altogether is 3!*1 ways. Tell my why my understanding is wrong

[uttam.albela]

Out of 6 places, select 3 places. No of ways = 6C3
No. of ways to Arrange these letters DBL in selected 3 places = 3!
Now we are left with 3 places where we need to put OUE in this same order.
No of ways to do this = 1

So total number of ways = 6C3 * 3! * 1 = 6!/3! = 120

Let me know if any part of it is not clear to you.

Thanks for the post! I'm still not clear why we are doing 6C3 here. Otherwise i'm clear abt rest of the steps.

See we have 6 places and we need to select 3 place to arrange 3 non-vowels. So number of ways to choose 3 places out of 6 is 6C3. Once we have 3 seats for 3 non-vowels, we can arrange then in 3! ways.

My method:
Consider vowels first and this can be arranged in only 1 way. -- I agree vowels can be arranged in only one way, but only after we have selected the 3 places for vowels. 3 seats for vowels can be selected from 6 seats available in 6C3 ways. Is it clear to you now?
Then remaining three places can be arranged with 3 remaining letters in 3! ways. - This is correct.
Altogether is 3!*1 ways. Tell my why my understanding is wrong

[papgust]

Hi All,

I have a basic doubt with this type of question again. If you haven't read the question posted in this topic, please take a few min to read and then read my question below.

Consider this question which is very similar according to me. Please bear with me if my question is not very clear to you as i re-phrased the question myself! I hope you understand what i'm trying to ask.

In how many ways can you re-arrange the letters in the word "DOUBLE" such that first letter must start with "D" and last letter must end with "E"?

For this qn, the answer will be 4! because because we are not considering D and E as we leave as it is and re-arranging the remaining 4 letters (I saw a similar kind of solution in a book)

My question here is why are we not multiplying 4! with 6C4. For me, both these problems look similar and also ask a similar question. Kindly clarify my doubt!! Permutations and Combinations are sometimes driving me crazy!!

[life is a test]

We have 6 letters in DOUBLE, so there are 6 places to be filled by these 6 letters..

Let us first think of arranging DBL.

Out of 6 places, select 3 places. No of ways = 6C3
No. of ways to Arrange these letters DBL in selected 3 places = 3!
Now we are left with 3 places where we need to put OUE in this same order.
No of ways to do this = 1

uttam.albela - to me 6C3 indicates that there are 6 options open to us in order to slot the 3 non vowel letter into but 3 places already occupied with stated vowels so that means we only have 3 options not 6?
can you see where i am going wrong? thanks.

[uttam.albela]

Let us just rephrase this question. There are 6 chairs. 6 Persons are to be seated. 3 of them named A, B, and C want to be in the same order. But other people may or may not seat between them.

We can plan for the sitting in two ways.
First make the other 3. and then A,B and C.
Solution:
6 empty chairs and We plan to seat 3 others and then later worry about A,B and C.
Choose 3 out of 6 chairs and then make 3 others seat.
So no. of ways to choose = 6C3
No of ways to seat them = 3!
Now left with 3 seats for A,B and C. They must see in this order. So it can be achieved in one way.


Second approach, make A,B and C seat and then the rest.
Here also first select 3 chairs from 6 and make A,B and C seat. Then we should worry about seating others.
Choose 3 out of 6 chairs for A,B and C to seat. 6C3
Only one way to seat these 3 fussy people.

Now 3 seats available for 3 non-fussy people. So 3! ways

[papgust]

So you are saying that both approaches lead to same answer i.e. 6C3 * 3!.

But consider this qn,

In how many ways can you re-arrange the letters in the word "DOUBLE" such that first letter must start with "D" and last letter must end with "E"?

This answer is 4!. My qn is why are we not multiplying 6C4 with 4! here. Even in this qn, we can say that first we need to keep D and E and then re-arrange other 4 letters or first arrange 4 letters and then take D and E. In either case, according to you, we must multiply by 6C4 with 4!. But OA is only 4!.

Hope i'm making sense!

[uttam.albela]

So you are saying that both approaches lead to same answer i.e. 6C3 * 3!.

But consider this qn,

In how many ways can you re-arrange the letters in the word "DOUBLE" such that first letter must start with "D" and last letter must end with "E"?

This answer is 4!. My qn is why are we not multiplying 6C4 with 4! here. Even in this qn, we can say that first we need to keep D and E and then re-arrange other 4 letters or first arrange 4 letters and then take D and E. In either case, according to you, we must multiply by 6C4 with 4!. But OA is only 4!.

Hope i'm making sense!

No Dear,

let me explain this question to you.

There are 6 persons named D, O, U, B, L, and E.
There are 6 chairs also. But D, E are kids adamant to occupy the first and last seat and they capture it respectively.

So the actual question becomes 4 chairs and 4 non-fussy adults to be seated.
So here we can say select 4 middle chairs from available 4 chairs. So it is 4C4 = 1.
Then number of ways to make 4 persons to make them seat on these 4 places are 4!.

So answer is 4!.