The number of arrangements of the letters of the word pankajpandey such that vowels come together
Find the number of permutations of the letters of the word ‘PENDULAM', such that vowels are never together. Show (Approach 1)I know one solution is to find the permutation that vowels are together (ie; 6!) and subtract it from total number of permutations for the word (ie; 8!). Approach 2: I came across another interesting approach like shown below : Consonants in the word = $5$ ie;{P,N,D,L,M} Ways of arranging $5$ letters = $5!$ Consider a case like PNDLM and we insert positions that satisfy the condition of vowels not being together like _P_N_D_L_M_. So the vowels can be in any of the $6$ dashed positions (which is a equivalent to choosing $3$ out of $6$ positions ie; $6C3$) and vowels for chosen $3$ positions can be arranged among themselves in $3!$ ways. So the number of permutations should be : $6C3 \cdot 3! \cdot 5! = 14400$ (using counting principle) whereas solution by Approach 1 would be $= 8! - 6! = 39600$. Can someone point out where Approach 2 went wrong. I would like to use Approach 2 so that I can solve cases like
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So,my book gives the answer $144$. I tried to do it this way-$$\text{_ E _ A _ O _ }$$However I cannot fill the blanks in $4P3$ $\times3!=144$ ways with the remaining letters as there might be words like $\text{CEAMOT}$ in which two vowels are adjacent. So,I tried to separate two cases like this-$$\text{_ E _ A _ O and E _ A _ O _} $$ which can be arranged in total $(3!\times3!)\times2=72$ ways. Am I wrong? asked Oct 18, 2017 at 5:14
$\endgroup$ 5 $\begingroup$ You left out two cases which are of the form of $$A**E*O$$ $$A*E**O$$ Hence $$(3!\times 3!)\times 4$$ answered Oct 18, 2017 at 5:17
Siong Thye GohSiong Thye Goh 145k20 gold badges83 silver badges146 bronze badges $\endgroup$ $\begingroup$ Here is an approach for separating the vowels which eliminates the need to consider cases. We will arrange three blue balls and three green balls in a row, with the green balls separated. Line up the three blue balls. This creates four spaces. $$\square \color{blue}{b} \square \color{blue}{b} \square \color{blue}{b}\square$$ To separate the three green balls, choose three of these four spaces in which to place a green ball. For instance, if we choose the first, second, and fourth spaces, we obtain the arrangement $$\color{green}{g}\color{blue}{b}\color{green}{g}\color{blue}{b b}\color{green}{g}$$ Arrange the three consonants in the three positions occupied by the blue balls and the three vowels in the three positions occupied by the green balls. Hence, there are $$\binom{4}{3}3!3! = 144$$ arrangements of the six letters in which the three vowels are separated. answered Oct 18, 2017 at 11:03
N. F. TaussigN. F. Taussig 68k13 gold badges52 silver badges70 bronze badges $\endgroup$ 3
Post your comments here:Name *: Email : (optional) » Your comments will be displayed only after manual approval. How many arrangements are there of the word vowels?=3×2×4×3×2=144.
How many arrangements of the letter of word examination be made such that vowels come together?So, there are 18,14,400 ways in which the word 'EXAMINATION' can be arranged by keeping the first letter as 'M'. Since all the vowels have to come together we are going to treat them as single letters. We treat them as EAIAIO single objects.
How many ways word arrange can be arranged in which vowels are not together?number of arrangements in which the vowels do not come together =5040−1440=3600 ways.
How many words can be formed such that all vowels are together?Number of words each having vowels together = 24 x 2 = 48 ways. Total number of words formed by using all the letters of the given words = 5! = 5 x 4 x 3 x 2 x 1 = 120. Number of words each having vowels never together = 120-48 = 72.
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