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Find the number of ways in which $A,B,C,a,b,c$ can be arranged so that the capital letters occur before the corresponding small letters.

Actually this question can be generalised to further $2n$ elements such that the $2n$ elements are separated into $2$ subgroups each containing $n$ elements in such a way that for each element in the first group $(A_i)$ there is an element in the second group $(a_i)$ which always occurs after the capital letter occurs.

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At present, I have 2 approaches for this problem.

Approach 1

For the trivial case, i.e., the first case, the number of ways such arrangement exists is ${6 \choose 2}\times{4 \choose 2}\times1$

I think this should be quite clear why, however this happens as follows. We choose any capital letter-small letter pair. These two elements can be placed in ${6\choose 2}$ in 6 spaces. Here, we are not overcounting the number of cases as, we are just choosing $2$ places and we do not specify the order in which we are arranging the elements. Continuing this process, we can have the complete solutions.

After the example, it should be quite easy to generalise the process. So, one can tell how many arrangements exist for $2n$ elements. This is supposed to be in ${2n\choose 2}\times{2n-2 \choose 2}\times\ldots\times 1$.

However, in my opinion, this method doesn't have much elegance. So, my preferred way would be the second approach.

Approach 2

To understand the approach in the simple way, we first treat the first case.

We arrange the $6$ elements in $6!$ ways. But, this way produces many cases of overcounting, quite obviously, but, we have a way to counter that. For every arrangement of a pair, there are $2$ cases, one in which the small comes after the capital letter, and the one in which the capital comes after the small letter, so, we divide the number of arrangements by $2^{\text{number of pairs}}$, i.e., in this case $2^3$.

So, now, it is quite easy to generalise this case, the number of ways, such an arrangement is possible is $\dfrac{(2n)!}{2^n}$.

I consider this problem to be open still and I am open to new approaches.

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    $\begingroup$ I like your second approach $\endgroup$ – Hagen von Eitzen May 6 '14 at 12:13
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    $\begingroup$ @HagenvonEitzen Honored that you like it...thank you... $\endgroup$ – Hawk May 6 '14 at 12:17
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Simple: There are $3!$ ways of reordering the upper case letters, ditto for lowercase. As the order is fixed, it is just $3! \cdot 3!$

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  • $\begingroup$ I think I meant both the cases, please see my solution. $\endgroup$ – Hawk May 6 '14 at 12:07
  • $\begingroup$ @Hawk, then your question isn't clear enough. $\endgroup$ – vonbrand May 6 '14 at 12:09
  • $\begingroup$ $(+1)$ Even though it is not really an answer, this was initially how I myself understood the question :) $\endgroup$ – Mr Pie Jun 1 '18 at 4:10

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