In how many ways can the letters of the word MULTIPLE be arranged in the following cases -?
(i) without changing order of the vowels,
(ii) keeping position of the vowels fixed
(iii) without changing the relative order of vowels and consonants.
- Pramod KumarLv 76 months agoFavourite answer
Here is the solution to your problem -
- PuzzlingLv 76 months ago
There are 8 letters, but L appears 2 times. So without restriction, there are 8!/2! = 20,160 ways to arrange the letters.
The problem with this question is I'm not 100% sure of their intended meaning for each of the restrictions. I'll add my assumptions and you can check them against what you think they mean.
If I'm understanding the question, they want any arrangement, but U, I and E must be in that order somewhere in the arrangement.
In any of the possible arrangements (without restriction), the vowels could appear equally in one of 3! = 6 orders. So if we divide by 6, we'll have the number of ways that U, I, E will still be in that order.
8! / (2! 3!)
= 40,320 / 12
= 3,360 ways
Now they want the vowels to be in their current positions (2, 5 and 8) but they could be rearranged (3! = 6 ways). Then the consonants could placed in the remaining 5 positions in 5! / 2! = 60 ways (don't forget the repeated Ls)
3! * 5!
= 6 * 60
= 360 ways
Here the vowels must be U,I,E and the consonants M,L,T,P,L
Just imagine 8 positions and pick the places for the 3 vowels. There is 1 order for them in those chosen positions and 1 order for the consonants that go in the other positions. So we just need to calculate "8 choose 3". Equivalently we could pick the 5 positions for the consonants then place the vowels. The value of "8 choose 5" gives the same result.
C(8,3) = C(8,5)
= (8 * 7 * 6) / (3 * 2 * 1)
= 56 ways