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Pr0bability
 2 years ago
Best ResponseYou've already chosen the best response.0It's a theorem that comes from combinatorics. Binomial is a polynomial with two (bi) terms. It basically says this: \[\left( a + b \right)^{n}=\sum_{m=0}^{n} \left( \left(\begin{matrix} n \\ m \end{matrix}\right) a ^{nm} b ^{m}\right)\] So for example: \[\left( a + b \right)^{3}=\left(\begin{matrix}3 \\ 0\end{matrix}\right) a ^{3} b ^{0} + \left(\begin{matrix}3 \\ 1\end{matrix}\right) a ^{2} b ^{1} + \left(\begin{matrix}3 \\ 2\end{matrix}\right) a ^{1} b ^{2} +\left(\begin{matrix}3 \\ 3\end{matrix}\right) a ^{0} b ^{3} \] Which is nothing else then: \[a ^{3}+3a ^{2}b+3ab ^{2}+b ^{3}\] It can be proven with mathematical induction.

jahnavi
 2 years ago
Best ResponseYou've already chosen the best response.0I am not able to understand,can explain in a more simpler way?

MattBenjamins
 2 years ago
Best ResponseYou've already chosen the best response.0A binomial is (as explained above) an expression that has two terms. The general form is: \[(a+b)^{n}\] Where a is the first term, b the second and n is the power to which the whole gets raised. If n is low expanding the expression is easily done. For example: \[(x+2)^2 = (x+2)(x+2) = x^2+2x+2x+4 = x^2+4x+4\] When n is high though, this kind of expansion is a lot of work and there is an easier way to do it which is called the Binomial Theorem: \[(a+b)^n = \sum_{m=0}^{n}\left(\begin{matrix}n \\ m\end{matrix}\right)a^{n1}b^m\] The first "scary bit" is what's called the Sigma Notation. In it's geneal form: \[\sum_{k=1}^{n}a _{k}\] This is just a simpler way to write the summation of many terms. For example: \[\sum_{k=1}^{4}2k=2+4+6+8=20\] In this example there are four terms in which some operation is aplied to k (in this case multiplication by two) and in the first term k=1. For each following term 1 gets added to k. The second thing is: \[\left(\begin{matrix}n \\ m\end{matrix}\right)\] Read: 'n choose m'. This means: the number of m sized sets you can take out of an n number of objects when order doesn't matter. Don't worry if that sounds confusing, to use the binomial theorem you only need to be able to calculate it which is done as follows: \[\left(\begin{matrix}n \\m\end{matrix}\right)=\frac{ n! }{ m!(nm)! }\] Where n! (read n factorial) means:\[1\times2\times3\times...\times(n2)\times(n1)\times(n )\]For example: \[7! = 1\times2\times3\times4\times5\times6\times7\times=5040\] So for example: \[\left(\begin{matrix}4 \\ 2\end{matrix}\right)=\frac{ 4! }{ (2!)\times(2!) }=\frac{ 24 }{4 }=6\] Note that when m=0: \[\left(\begin{matrix}n \\0\end{matrix}\right)=1\] Hopefully this explanation can help you understand the example in the firts reply.
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