Binomial Co-efficients

 C(n,k) qk ----> Probability -> x (Y) Total trials k <-- Qk k-> 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 total:2n 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20

ΣnC= 1 (k=0)

ΣnC= 1 + n (k=0 to k=1)

ΣnC= 1 + n(n+1)/2 (k=0 to k=2)

ΣnC= 1 + n(n+1)/2 +( n(n-1)(n-2)/2)  - Σn-->3 to n[(n-1)(n-2) + (n-2)(n-3) +...... (n-n)(n-n+1)]  (k=0 to k=3)

ΣnC = 1 + n(n+1)/2  + [n(n-1)(n-2)/2] - Σn-->3 to n[(n-1)(n-2) + (n-2)(n-3) +...... (n-n)(n-n+1)] +1/2 [1*(n-2)(n-3)+2*(n-3)(n-4) +

3*(n-4)(n-5) + .... +(n-3)*(n-(n-2))(n-(n-1)) ] (k=0 to k=4 and (n-3) is the last term co-efficient)

Qk  is the ratio of frequency for k and frequency for k-1 i.e Qk =((n+1-k)/k)*(x/y) =qk (x/y)

where qk  = (n+1-k)/k

Property of qk and Qk       (excluding behavior at 0≤k<1)

*Qk  monotonically decreases as k increases, n remaining constant. Qk (maximum) = n*(x/y)      for k=1

Qk (minimum) =(1/n)*(x/y)   for k=n

*qk  monotonically decreases as k increases, n remaining constant.  qk (maximum) = n      for k=1 (qk maximum value being ≥1)

qk (minimum) =(1/n)  for k=n (qk minimum value being  <1)

qk  =1 when k=(n+1)/2  i.e. when n is an odd number since k has to be an integer.

Qk  =1 when k=(n+1)/(z+1)  where z = (y/x) and k has to be an integer.

* If  we define a function  f (x) such that Y = f (x) = (n+1-x)/x  (do not confuse Y with small y) , then

Y =(n+1)/x  - 1. Define another function  z =f (Y) = Y + 1 , then  z =f1(x) = (n+1)/x. Since n+1 is a constant say A, z = A/x or zx=A

This is the Equation of a Equilateral or Rectangular Hyperbola with asymptotes taken as the co-ordinate axis and is of the form xy=a2/2 = constant. The coordinates of the vertices are on the bisector of the first and third quadrant and the first and second coordinate coincide, that is to say, x = z. Also, Point A belongs to the curve of the hyperbola.

Vertices lie in 1st,3rd quadrant and are (√A,√A) and (-√A,-√A)

Length of each Semi-Axis (here a=b) a, b is √(2A). Length of Semi -axis c is √(a2+b2) =√(a2+a2) = √(2a2) =√(2.2A) =2√A

Focus is (√(2A),√(2A) ) , (-√(2A),-√(2A) ). Eccentricity=c/a= √2 (Eccentricity of hyperbola is always greater than 1)