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This thing is confusing me for a long time and hence I am asking it here.

We have a given set of numbers $$ X = \{1,2,3,4,5\} $$ and we want to form a $4$- digit number of the form $aaac$(this is just symbolic; I want all the arrangements of this form for e.g. $aaca$ etc). (I know that such questions' answers do exist on MSE, but my doubt/ confusion is pertinent to something extraneous to such questions.)

To make such a number the first approach that we can induce in our solution is that of using simple combination technique and choose two numbers out of the set of $5$ numbers. This can be done in the following way:

$$\binom{5}{2} \cdot \frac {4!}{3!}=40 $$

That seems fine as far as the combination formula is confirmed. Let us now take a naive approach and pretend as if we don't know the combination formula.

Here we use the gap method: Let us make one of our configuration as: $$\fbox{a} \fbox{a} \fbox{a} \fbox{c}$$ Here $a$ can be chosed in $5$ ways and $c$ can be chosen in $4$ ways Total ways to make $aaac$ configuration $=5 \cdot 1\cdot 1\cdot 4 = 20$ ways

Every such configuration now needs to be permuted since we want $aaca$ etc. also. To permute such a configuration, we can write: $$ \frac {4!}{3!} \cdot 20 = 80$$ which is exactly double that of the answer obtained from the combination formula (which makes be believe that either I have double counted things or I have snubbed some of the arrangements).

Now let us try to think about some cases manually:

Case $1$: When $a=1$ The cases are: $$\{1112\}, \{1121\},\{1211\},\{2111\},\{1113\},\{1131\},\{1311\},\{3111\},\{1114\},\{1141\},\{1411\},\{4111\},\{1115\},\{1151\},\{1511\},\{5111\}$$ Thus there are $16$ configurations regarding $1$ only which means that for all the whole set, there are $16 \cdot 5 =80$ configurations.

This proves that the problem is with my usage of the combination formula. Can anybody tell me how my usage of combination formula comes out to be fallacious?

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The problem here is that $a$ and $c$ have two different roles. Therefore simply using $\binom{5}{2}$ will not work, as it denotes the number of ways of picking two elements out of five, but without distinguishing between them. What you can do is $$\binom{5}{2}\cdot2\cdot4 = 80\ ,$$ where we piked two numbers in $\{1,2,3,4,5\}$ using $\binom{5}{2}$, then we choose which one to assign to $a$ and which one to $c$ (whence the factor $2$), and finally we choose where to put the $c$ (the factor $4$).

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    $\begingroup$ +1! We have almost the exact same formatting for the final expression :) $\endgroup$ – Zubin Mukerjee Jun 5 '17 at 21:25
  • $\begingroup$ And if you would generalize, I would write this as $\binom{5}{2} \cdot 2! \cdot \binom{4}{1}$. $\endgroup$ – CompuChip Jun 6 '17 at 7:21
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There are $\binom{5}{2}$ ways to choose two numbers from the set of $5$. However, there are addditionally $2$ ways to choose which of those two numbers will be repeated three times. Given this, there are finally $4$ ways to choose which of the four positions the single non-repeated number will be in. This gives

$$\binom{5}{2} \cdot 2 \cdot 4 = \boxed{80\,}$$

This matches your answer using the other method.

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You have to choose $2$ numbers. Which one number is repeated three times like aaac etc. Possible ways; $$\binom{5}{2}$$

Positions; $4$

Numbers to be chosen; $2$

$$\binom{5}{2} \cdot 4 \cdot 2 = 80$$

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jun 7 '17 at 0:09
  • $\begingroup$ okay. i try to read this tutorial. and thanks $\endgroup$ – ABDUL KHALIQ Jun 7 '17 at 9:42

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