[like_dreams]

HI,
please help me with the logic of following questions -

How many words can be formed with the letters of word 'MATHEMATICS' IF -

1. The order of vowels do not change
2. No two vowels occupy consecutive place

[thakarsagar]

1) When the order of vowels remains unchanged...
Total no of permutations = 11! / (2! * 2! * 4!) = 415800
All the 4 vowels can be considered to be alike objects...

2) When all vowels have to be together...
Permutations = 8! * 12/ (2! * 2! ) = 120960.
Here 12 is the no of ways in which these 4 vowels can be permuted together.

3) When no 2 vowels can be together.
Total no of permutations = 11! / (2! * 2! *2!)

Permutations when AA occur together = 10! / (2! * 2!).
Permutations when EA occur together = 10! * 2!/ (2! * 2!).
Permutations when IA occur together = 10! * 2!/ (2! * 2! *2!).
Permutations when EI occur together = 10! * 2!/ (2! * 2! *2!).

Subtract these cases.

[made_for_iims]

Originally Posted by thakarsagar

1) When the order of vowels remains unchanged...
Total no of permutations = 11! / (2! * 2! * 4!) = 415800
All the 4 vowels can be considered to be alike objects...

2) When all vowels have to be together...
Permutations = 8! * 12/ (2! * 2! ) = 120960.
Here 12 is the no of ways in which these 4 vowels can be permuted together.

3) When no 2 vowels can be together.
Total no of permutations = 11! / (2! * 2! *2!)

Permutations when AA occur together = 10! / (2! * 2!).
Permutations when EA occur together = 10! * 2!/ (2! * 2!).
Permutations when IA occur together = 10! * 2!/ (2! * 2! *2!).
Permutations when EI occur together = 10! * 2!/ (2! * 2! *2!).

Subtract these cases.

In your second answer.............
You have considerd these fours sets as distincts....but they will not be....
So you should add to the total (or subtract from this)....the cases where all vowels are together....coz u have counted them more than once......

[Apprentice]

Originally Posted by like_dreams

HI,
please help me with the logic of following questions -

How many words can be formed with the letters of word 'MATHEMATICS' IF -

1. The order of vowels do not change
2. No two vowels occupy consecutive place

2-Place all consonants in line, so now te spaces available for vowels are denoted by X...
X M X T X H X M X T X C X S X...

Terefore, 8 spaces available for 4 vowels(A, E, A, I)...Therefore no of possible permutations :8P4/2!....(2! bcos there are 2 As)...

And Permutations possible for consonanats:7!/(2! * 2!)...

hence total Permutations (8P4/2!) * 7!/(2! * 2!)....

Hope I have made it as lucid as possible...

[sujoygolan]

Originally Posted by thakarsagar

1) When the order of vowels remains unchanged...
Total no of permutations = 11! / (2! * 2! * 4!) = 415800
All the 4 vowels can be considered to be alike objects...

2) When all vowels have to be together...
Permutations = 8! * 12/ (2! * 2! ) = 120960.
Here 12 is the no of ways in which these 4 vowels can be permuted together.

3) When no 2 vowels can be together.
Total no of permutations = 11! / (2! * 2! *2!)

Permutations when AA occur together = 10! / (2! * 2!).
Permutations when EA occur together = 10! * 2!/ (2! * 2!).
Permutations when IA occur together = 10! * 2!/ (2! * 2! *2!).
Permutations when EI occur together = 10! * 2!/ (2! * 2! *2!).

Subtract these cases.

Can somebody clarify the above answer to the 1st question:
Should it not be 8!/(2!*2!) if all vowels are considered as one object in the same order ?

[pendyal]

Originally Posted by sujoygolan

Can somebody clarify the above answer to the 1st question:
Should it not be 8!/(2!*2!) if all vowels are considered as one object in the same order ?

yes...u r right....it will be 8!/(2!*2!)...since we consider all the vowels in the same order as one alphabet say X and then we have to permute M,T,H,M,T,C,S AND X which can be done as u said in 8!/(2!*2!)

[thakarsagar]

Originally Posted by pendyal

yes...u r right....it will be 8!/(2!*2!)...since we consider all the vowels in the same order as one alphabet say X and then we have to permute M,T,H,M,T,C,S AND X which can be done as u said in 8!/(2!*2!)

No...they cannot be considered as one object... beause they may occur at different places while their ordering may remain the same...

when you are considering 8! , u assume they occur together...thats not the case here...

But when you consider them as alike , u r ensuring a single order but may be at different places. That was one of the reason why i gave (2) solution though it was'nt mentioned in the question...

[pendyal]

yeah....i overlooked that fact while doing the problem....thanks for pointing it out......

Originally Posted by thakarsagar

No...they cannot be considered as one object... beause they may occur at different places while their ordering may remain the same...

when you are considering 8! , u assume they occur together...thats not the case here...

But when you consider them as alike , u r ensuring a single order but may be at different places. That was one of the reason why i gave (2) solution though it was'nt mentioned in the question...

[catdog]

Originally Posted by thakarsagar

1) When the order of vowels remains unchanged...
Total no of permutations = 11! / (2! * 2! * 4!) = 415800
All the 4 vowels can be considered to be alike objects...

2) When all vowels have to be together...
Permutations = 8! * 12/ (2! * 2! ) = 120960.
Here 12 is the no of ways in which these 4 vowels can be permuted together.

3) When no 2 vowels can be together.
Total no of permutations = 11! / (2! * 2! *2!)

Permutations when AA occur together = 10! / (2! * 2!).
Permutations when EA occur together = 10! * 2!/ (2! * 2!).
Permutations when IA occur together = 10! * 2!/ (2! * 2! *2!).
Permutations when EI occur together = 10! * 2!/ (2! * 2! *2!).

Subtract these cases.

Hi for 1st problem answer is rt but interpretation is wrong ...we divide it by 4! not because they are like objects because once in every sixteen arrangement the order is preserved for ex DFGH jMNk ..Now if u arrange this word in all possible manner 1/2 times j will b ahead of k and 1/2 times k wikk b ahead of j ....

[akankshak]

Just as an elaboration to the first question

1. As a formula the number of ways in which n things can be arranged is n!. Now the word MATHEMATICS has 11 numbers hence they can be arranged in 11! ways. As a formula the number of ways in which n things can be arranged in which r are alike is n!/r! where r is the number of alike things . Here the letter M is repeated twice and letter T is repeated twice . Also as the order of the vowels remains constant we can consider them like four alike things, so only one ordering is possible

Hence permutation = 11!/2!*2!*4!

Originally Posted by like_dreams

HI,
please help me with the logic of following questions -

How many words can be formed with the letters of word 'MATHEMATICS' IF -

1. The order of vowels do not change
2. No two vowels occupy consecutive place