For a particular problem, I have arranged the letters A and B twelve by twelve : AAAAAAAAAAAA, ABBAABBAAABA, and so on. Order matters: AB is not equal to BA. Altogether there are 2 raised to 12 or 4096 pemutations giving all possible combinations of A and B, twelve by twelve. Now amongst the 4096 lines of all the combinations, I want to select just the lines which have A six times, like for instance AABBBBBAAABA. How many such lines are there amongst the 4096? I tried the formula 4096!/(4096-6)! but I am not sure that this does the job, since it does not specifically choose lines with six times A.

Thanks for the answer. It is important to me.

There are \left( \array 12\\ 6 \array \right) ways to place the 6 A's in the twelve letter positions. Thus:

\left( \array 12\\ 6 \array \right) = \frac {12!/6!}{6!} = \frac {12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1 } =924

In general the number of ways that 'k' A's can be placed in a string of 'n' letter positions is:


\left( \array n\\ k \array \right) = \frac {n (n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...1} = \frac {n!}{k!(n-k)!}

There are \left( \array 12\\ 6 \array \right) ways to place the 6 A's in the twelve letter positions. Thus:

\left( \array 12\\ 6 \array \right) = \frac {12!/6!}{6!} = \frac {12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1 } =924

In general the number of ways that 'k' A's can be placed in a string of 'n' letter positions is:


\left( \array n\\ k \array \right) = \frac {n (n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...1} = \frac {n!}{k!(n-k)!}


Thanks a million for your answer. I suppose that this is independent of the total sample of 4096, all the ways you can arrange A and B, twelve by twelve. Since you are very knowledgeable, I want to ask you a further question in order to solve my problem completely. Now that I have reduced my sample to 954, keeping just the lines with six A's, I want to reduce it still further by just keeping the lines where three A's come in a row, side by side: AAA. How manu such lines will there be amongst the 954? Sorry for this further bother.

Thanks a million for your answer. I suppose that this is independent of the total sample of 4096, all the ways you can arrange A and B, twelve by twelve. Since you are very knowledgeable, I want to ask you a further question in order to solve my problem completely. Now that I have reduced my sample to 954, keeping just the lines with six A's, I want to reduce it still further by just keeping the lines where three A's come in a row, side by side: AAA. How manu such lines will there be amongst the 954? Sorry for this further bother.

Sorry, in the above, please read 924 instead of 954. Thank you.

[QUOTE=Sabarkantha;3583162]Sorry, in the above, please read 924 instead of 954. Thank you very much.

To clarify your question - I assume you mean that if you get 4 or more A's in a row it's counted as a fail, and if you get two groups of 3 A's each separated by at least one B it's a success.

Start by considering how the three A's are arranged relative to B's:
Case 1. You can have the three A's as the first three letters followed by a B, and you don't care how the remaining 8 letters are arranged:: AAABxxxxxxxx.
Case 2. The first N letters can be anything, followed by the pattern BAAAB, and then the remaining 7-N letters can be anything; for example if N=2: xxBAAABxxxxx. This is valid for N=0 to N=7.
Case 3. The first 8 letters can be anything, followed by BAAA for the last four letters: xxxxxxxxBAAA.

For case 1 the number of ways that the last 8 letters can be arranged, given that they consist of 3 A's and 5 B's is:


\left( \array 8 \\ 3 \array \right) = \frac {8 \cdot 7 \cdot 6}{3!} = 56.


For case 2 the number of ways that the remaining 7 letters can be arranged, given that there are 3 A's and 4 B's, is:


\left( \array 7 \\ 3 \array \right) = \frac {7 \cdot 6 \cdot 5}{3!} = 35.


Note that case 2 can occur in 8 different ways, since the starting B can be in position 1, 2, 3, 4, 5, 6, 7, or 8. Thus the total number of ways that case 2 can occur is 8 x 35 = 280.

Finally for case 3 the math is the same as for case 1, so there are 56 ways it can occur.

Thus the total number of arrangements that yield 3 A's in a row, given that you have 6 A's and 6 B's, is 2x56+280 = 392.

To clarify your question - I assume you mean that if you get 4 or more A's in a row it's counted as a fail, and if you get two groups of 3 A's each separated by at least one B it's a success.

Start by considering how the three A's are arranged relative to B's:
Case 1. You can have the three A's as the first three letters followed by a B, and you don't care how the remaining 8 letters are arranged:: AAABxxxxxxxx.
Case 2. The first N letters can be anything, followed by the pattern BAAAB, and then the remaining 7-N letters can be anything; for example if N=2: xxBAAABxxxxx. This is valid for N=0 to N=7.
Case 3. The first 8 letters can be anything, followed by BAAA for the last four letters: xxxxxxxxBAAA.

For case 1 the number of ways that the last 8 letters can be arranged, given that they consist of 3 A's and 5 B's is:


\left( \array 8 \\ 3 \array \right) = \frac {8 \cdot 7 \cdot 6}{3!} = 56.


For case 2 the number of ways that the remaining 7 letters can be arranged, given that there are 3 A's and 4 B's, is:


\left( \array 7 \\ 3 \array \right) = \frac {7 \cdot 6 \cdot 5}{3!} = 35.


Note that case 2 can occur in 8 different ways, since the starting B can be in position 1, 2, 3, 4, 5, 6, 7, or 8. Thus the total number of ways that case 2 can occur is 8 x 35 = 280.

Finally for case 3 the math is the same as for case 1, so there are 56 ways it can occur.

Thus the total number of arrangements that yield 3 A's in a row, given that you have 6 A's and 6 B's, is 2x56+280 = 392.

Thanks once again for the perfectly clear answer. You have understood the problem well. I think that I can work things out myself now.