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4 Answers

Best answer
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METHOD: if we place the vowels first.

There are 5 vowels in PERMUTATION, WHICH has 11 letters.

We can select positions for the five vowels in C(11,5) ways.There is only one way to arrange the vowels in those positions in alphabetical order.

There are C(6,2) ways to place the T's in two of the remaining 6 positions.

There are 4! ways to arrange the P,M,N and R in the remaining 4 positions. Hence the number of distinguishable arrangements of PERMUTATION in which the vowels appear in alphabetical order is

C(11,5)*C(6,2)*4!

= {11!/(5!*6!)}*{6!/ (2!*41)} *4!

= {11!/(5!*2!)}

=166320 WAYS

for better explanation:-

http://math.stackexchange.com/questions/1448684/ways-to-arrange-the-letters-in-bookkeeper-if-vowels-must-appear-in-alphabeti

 

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Here we have total 11 characters where T=2 and vowels are (a,e,i,o,u) = 5.

Total no of words = 11 ! / 2! but all 5 vowels can not be permuted, since it has to maintain alphabetic order.

so ans -  (11 !) / ( 5 ! * 2 !) = 1,66,320

edited by
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Answer will be 6! /2!

As, there are 5 different vowel they arrange in 1 way

and 6 consonant and 1 repeating . So, they could arrange among themselves in 6!/2! ways
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One other approach to solve this problem will be fix positions for 5 vowels in alphabetical order. Now there are 6 positions to place constant. (i.e. C V C V C V C V Where C is constant and V is vowel). Now find the number of permutations of non-negative integer solutions for equation X1 + X2 + X3+ X4+ X5+ X6 = 6 since there are 6 places to be filled(X1, X2..... denotes number of constants at position 1,2.... respectively) and and the sum of elements at those 6 places will be 6(since there are 6 constants to be placed). So r = 6 and n = 6 here. Putting  it in formula (n + r - 1) P (r - 1), i.e. 11 P 5 = 332640. Since two constants are same removing their permutation final answer is 332640/2 = 166320

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