papgust wrote: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
Thanks for the post! I'm still not clear why we are doing 6C3 here. Otherwise i'm clear abt rest of the steps.uttam.albela wrote: 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 wrote:Thanks for the post! I'm still not clear why we are doing 6C3 here. Otherwise i'm clear abt rest of the steps.uttam.albela wrote: 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.
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
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)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"?
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?uttam.albela wrote: 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
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!.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"?
No Dear,papgust wrote:So you are saying that both approaches lead to same answer i.e. 6C3 * 3!.
But consider this qn,
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!.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"?
Hope i'm making sense!
