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The question is how many ways the word INDIA can be re-written.

I N D I A

There are 5 letters in this word. The letter I is repeated twice, and this number doesn't make any sense if there positions are interchanged.

So,basically the number of ways INDIA can be arrange consider there are two types of I --- the answer is factorial 5. However the letter 'I' is repeated twice. Thus, we should subtract the number of ways the the two letter I is occurred. Thus, we should subtract it. However, we always divide it --- 5!/2. I want to understand, whey do we do so? Why do we divide by 2 here rather than doing subtraction? I couldn't get the logic here. Can someone please explain to me?

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The letter $I$ appears $2$ times.

So every arrangement that you have created appears $2!$ times.

Hence you need to divide the total number of arrangements that you have created by $2!$.


It just so happens that $2!=2$.

If the letter $I$ appeared $3$ times, then you would have to divide by $3!$ (i.e., by $6$).

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  • $\begingroup$ @dexterous_stranger: Thank you, feel free to accept it then :) $\endgroup$ – barak manos Sep 10 '14 at 14:50
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Your intuition to start with is good; we distinguish the two $I$s, lets say as $I_1$ and $I_2$. Then we take all possible permutations of the five (now distinct letters). If we consider any such permutation, say $N\ D\ I_1\ A\ I_2$, then there is always one other different permutation (in this case $N\ D\ I_2\ A\ I_1$) that will give me the same word when I forget the subscripts again. So counting every permutation of the five letters has counted every word twice, and so we have to divide by two to get the right answer - subtracting two would only make sense if the permutations had somehow produced two extra words.

As pointed out in barak's answer; if there were three letters the same, you would in fact have to divide by $6=3!$; given any permutation of your distunguished letters, you can re-order the three that are really supposed to be the same in any way, and you won't get a new word when you un-distinguish them again.

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  • $\begingroup$ Can you help me here - fence error problem - math.stackexchange.com/questions/923663/… $\endgroup$ – dexterous Sep 10 '14 at 14:52
  • $\begingroup$ @dexterous_stranger I think that question is too general for me to give a good answer to. $\endgroup$ – mdp Sep 10 '14 at 15:49
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Assume we have 4 balls , Number of ways of arranging 4 balls is Factorial of 4 ie 24. Now Suppose we have 1 Red Ball and 3 Black Balls ( ie, R B B B ) . Consider Black balls as B1 , B2 and B3 . Hence we have now (R B1 B2 B3) . We can have 4 ways of this permutation

R B1 B2 B3 | B1 R B2 B3 | B1 B2 R B3| B1 B2 B3 R Lets denote this Set as S1.

Now we Know that Number of ways of arrangements of black balls is 3 ! => 6 . Thus we will have 6 more sets like S1 but with different arrangements for black balls. ie Total we will have 24 such permutations ,Now as we initially knew that we will have 24 permutations of 4 balls . Also we now know Contribution of 3 same black balls in this 24 permutations ie 4 * 6 Now To eliminate these EXTRA Permutations , Answer would be 24 / 6 = > 4 ( Set S1) and not 26 - 6 => 18 Which is Wrong. Thanks.

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  • $\begingroup$ Please use Mathjax! $\endgroup$ – Arman Malekzadeh Jun 14 '17 at 7:21
  • $\begingroup$ @ArmanMalekzade I am new here , Thanks for the suggestion . $\endgroup$ – Archit Chauhan Jun 14 '17 at 9:39

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