Mathematics_Today_-_October_2016

z The number of ways in which m + n things can
be divided into two groups containing m and n
things respectively =()!+
!!

mn.
mn
z If n = m, the groups are equal, and in this case the

number of different ways of subdivision =

2

2

m

mm

!
!!!^
z If 2m things are to be divided equally between two

persons, then the number of divisions =

2 m

mm

!

!!

.

z The number of divisions of m + n + p things
into groups of m, n and p things respectively
=()!++
!!!

mn p

mn p

z If 3m things are divided into three equal groups,

then the number of divisions = ()!

!!!!

3

3

m

mmm

.

z If 3m things are to be divided among three persons,

then the number of divisions = ()!

!!!

3 m.

mmm

COMBINATION WITH REPETITION
z If there are a 1 objects of Ist kind, a 2 objects of 2nd
kind, a 3 objects of 3rd kind, ..., an objects of nth kind
and we want to choose r objects out of these under
the condition that at least one object of every kind
should be chosen.
Number of ways = Coefficient of xr in
(xx+++^22 ... x xxaa^12 )( +++... x)...(xx+++^2 ... xan)

z Number of ways in which r identical things can be
distributed among n persons when each person
can get zero or more things
= Coeff. of xr in (1 + x + x^2 + ... + xr)n

= Coeff. of xr in^1

1

−^1

−

⎛

⎝

⎜

⎞

⎠

⎟

x+

x

r n

= Coeff. of xr in [(1 – xr + 1 )n (1 – x)–n]

= Coeff. of xr in (1 – x)–n [leaving terms containing

powers of x greater than r]

= n + r – 1Cr

NUMBER OF INTEGRAL SOLUTIONS OF LINEAR
EQUATIONS AND INEQUATIONS
z Consider the equation
x 1 + x 2 + x 3 + x 4 + .... + xr = n ....(1)
where x 1 , x 2 , x 3 , x 4 , ..., xr and n are non-negative
integers.

This equation may be interpreted as that n identical

objects are to be divided into r groups where a group

may contain any number of objects. Therefore,

Total number of solutions of equation (1)

= Coefficient of xn in (x^0 + x^1 + ... + xn)r

= n + r – 1Cr or n + r – 1Cn (^) – 1.
z Total number of solution of the equation (1) when
x 1 , x 2 , x 3 ..., xr are natural number
= Coefficient of xn in (x^1 + x^2 + x^3 +.... + xn)r
= n – 1Cr – 1
z Consider the equation
x 1 + 2x 2 + 3x 3 + ... + qxr = n
where x 1 , x 2 , x 3 ....., xr and n are non-negative
integers then total number of solution of the above
equation is = Coefficient of xn in (1 + x + x^2 + ... )
(1 + x^2 + x^4 + ... ) (1 + x^3 + x^6 + ... ) ... (1 + xq + x^2 q

• ...)
  If zero is excluded, then the number of solutions
  of the above equation = Coefficient of xn in
  (x + x^2 + x^3 + ... ) (x^2 + x^4 + x^6 + ... ) (x^3 + x^6 + x^9 +
  ... ) ... (xq + x^2 q + ... )
  Note :
  (i) The coefficient of xr in (1 – x)–n is n + r – 1Cr.
  (ii) The coefficient of xr in (1 + x)–n is (–1)r n + r – 1 Cr.
  CIRCULAR PERMUTATIONS
  z If clockwise and anticlockwise orders are taken as
  different, number of circular arrangements of n
  different things taken all at a time = (n – 1)!
  If clockwise and anticlockwise orders are not taken
  as different, number of circular arrangements of n
  different things taken all at a time=^1 −
  2
  ()!n 1
  z If a circular arrangement can be flipped or turned
  upside down, number of circular arrangements
  of n different things taken all at a time=()!n−^1 ,
  2
  otherwise it is (n – 1)!
  z If positions in a circular arrangement are numbered,
  number of circular arrangements of n different
  things taken all at a time = n!, otherwise obviously
  (n – 1)!
  z Number of circular permutations of n different
  things taken r at a time if clockwise and anticlockwise
  orders are taken as different = nCr ⋅ (r – 1)!
  z Number of circular permutations of n different
  things taken r at a time if clockwise and
  anticlockwise orders are not taken as different
  =nCr()!r−^1
  2