- How to calculate total combinations for AABB and ABBB sets?
Although both belong to a much broad combination of N=2 and n=4 (AAAA, ABBA, BBBB ), where order matters and repetition is allowed, both can be rearranged in different ways: First one: AABB, BBAA,
- sequences and series - The Perfect Sharing Algorithm (ABBABAAB . . .
The algorithm is normally created by taking AB, then inverting each 2-state 'digit' and sticking it on the end (ABBA) You then take this entire sequence and repeat the process (ABBABAAB)
- 11 | abba, where a and b are the digits in a 4 digit number.
Truly lost here, I know abba could look anything like 1221 or even 9999 However how do I prove 11 divides all of the possiblities?
- Matrices - Conditions for $AB+BA=0$ - Mathematics Stack Exchange
There must be something missing since taking $B$ to be the zero matrix will work for any $A$
- How many $4$-digit palindromes are divisible by $3$?
Hint: in digits the number is $abba$ with $2 (a+b)$ divisible by $3$
- elementary number theory - Common factors for all palindromes . . .
For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of: $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are
- Is $AB+BA$ positive definite too if $A$ and $B$ are positive definite?
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- How many words of length $n$ can be generated from an alphabet of $k . . .
This is kind of late, but the correct approach would be to use binomial coefficients For instance, here we have words of length 4 and we're looking for a binary chain with 2 "1"s and 2 "0"s which would translate to C (4,2), which is 6 And with the number of different pairs of symbols you'd have k x (k-1) different combinations So the answer would be k x (k-1) x C (n,r) where k is your
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