For any set of n positive numbers ,
Arithmetic Mean >= Geometric Mean >=Harmonic mean
The equality occurs only when all the numbers are equal .
This can be used if the sum(or product ) of some numbers are given and the maximum(or minimum) value of the product(or sum ) is asked ...
Ex : xy=27
minimum value of 3x+4y ?
taking 3x and 4y to be two numbers and applying our rule
A.M : (3x+4y)/2
G.M : (12xy)^1/2=(4*3*27)^1/2 = 18
A.M >= GM
3x+4y >= 2*18
Hence the minimum value of (3x+4y) is 36
How we can find different pythagorus triplets.
The Formula is n+[n/(2n+1)]
hence N can take values from 1,2.....
hence the first mixed fraction becomes 1[1/3] = 4/3
One side of the pythagorus theorem is denominator which is 3 and other side is numerator which is 4, the hypotenuse is 4+1 = 5
Hence the triplet is 3,4,5.
Lets take n = 3
2n+1 = 7
Mixed fraction becomes = 24/7
the first side becomes 7, the other side is 24 and hypotenuse is 24+1 = 25
hence 7,24,25.
all the other remaining pythagorus triplet is multiples of the triplets given by the above formula.
Arithmetic Mean >= Geometric Mean >=Harmonic mean
The equality occurs only when all the numbers are equal .
This can be used if the sum(or product ) of some numbers are given and the maximum(or minimum) value of the product(or sum ) is asked ...
Ex : xy=27
minimum value of 3x+4y ?
taking 3x and 4y to be two numbers and applying our rule
A.M : (3x+4y)/2
G.M : (12xy)^1/2=(4*3*27)^1/2 = 18
A.M >= GM
3x+4y >= 2*18
Hence the minimum value of (3x+4y) is 36
______________________
2. Permutations and Combinations Concept :-
how to put 9 identical rings in 4 fingers.
concept= n+r-1Cr-1
this formula is used to distribute n identical things among r people.
if the 9 rings are named, a1,a2,a3....a7,a8,a9
we want to distribute them among 4 fingers, means we want to make four groups out of this 9 rings.
so if we arrange the 9 rings side by side.
_a1_a2_a3 _a4_ a5_ a6_ a7_a8_a9_
we need three separator to divide them in 4 groups. and we can put the separator at any of the black space above.
suppose I put first separator after a1, second after a4, 3rd after a6
so groups are
a1
a2,a3,a4
a5,a6
a7,a8,a9
that means I can say I have total 12 items ( 9+3) to arrange them selves.
that is 12!
but 9 rings are identical and 3 separators are also identical
so final answer shd be = 12!/9!*3
if we replace 9 with n,3 with r
we get
n+r-1Cr-1
2. how to distribute 9 different rings among 4 fingers.
just a single change rings are different, so we dont have to divide by 9!,
so answer = 12!/3!
general formula = n+r-1Pr-1
both of the above Q are of arrangement and distribution.
examples where this concept can be used.
1. distribute 1o chocolates aming 6 children such that no children is empty handed.
2. find whole number solutions for X+Y+Z = 22
3. find natural number solutions for X+Y+Z= 22
4. total number of terms in (a+b+c+d)^15
concept 3. total number of squares which can be made from size in N*N size square.
= 1^2 + 2^2 + 3^3 .....N^2
like for 2*2 square we can have total 5 squares, 4 square of 1*1, and the 2*2 square itself.
concept 4. total number of rectangles which can be made from N*N square.
= 1^3+2^3+3^3.....n^3
concept 5. A plane( restricted) is to be divided in N distinct parts, find the minimum number of lines to do so.
formula = > sigma X = N-1
X is the minimum number of lines.
suppose we want to divide plane in 16 distinct parts
sigma5 = 15
so answer is 5 lines.
if the 9 rings are named, a1,a2,a3....a7,a8,a9
we want to distribute them among 4 fingers, means we want to make four groups out of this 9 rings.
so if we arrange the 9 rings side by side.
_a1_a2_a3 _a4_ a5_ a6_ a7_a8_a9_
we need three separator to divide them in 4 groups. and we can put the separator at any of the black space above.
suppose I put first separator after a1, second after a4, 3rd after a6
so groups are
a1
a2,a3,a4
a5,a6
a7,a8,a9
that means I can say I have total 12 items ( 9+3) to arrange them selves.
that is 12!
but 9 rings are identical and 3 separators are also identical
so final answer shd be = 12!/9!*3
if we replace 9 with n,3 with r
we get
n+r-1Cr-1
2. how to distribute 9 different rings among 4 fingers.
just a single change rings are different, so we dont have to divide by 9!,
so answer = 12!/3!
general formula = n+r-1Pr-1
both of the above Q are of arrangement and distribution.
examples where this concept can be used.
1. distribute 1o chocolates aming 6 children such that no children is empty handed.
2. find whole number solutions for X+Y+Z = 22
3. find natural number solutions for X+Y+Z= 22
4. total number of terms in (a+b+c+d)^15
concept 3. total number of squares which can be made from size in N*N size square.
= 1^2 + 2^2 + 3^3 .....N^2
like for 2*2 square we can have total 5 squares, 4 square of 1*1, and the 2*2 square itself.
concept 4. total number of rectangles which can be made from N*N square.
= 1^3+2^3+3^3.....n^3
concept 5. A plane( restricted) is to be divided in N distinct parts, find the minimum number of lines to do so.
formula = > sigma X = N-1
X is the minimum number of lines.
suppose we want to divide plane in 16 distinct parts
sigma5 = 15
so answer is 5 lines.
______________________
3. NO. OF SQUARES AND RECTANGLES IN A CHESSBOARD
in a chessboard,there are 8*8 squares.
in a 2*2 chessboard, there are 5 squares (4 small aquares, 1 big square).
similarly in an n*n chessboard, there are 1^2 +2^2+....+n^2 squares.
so in a 8*8 chessboard, n=8
=> no.of squares = 1^2 +2^2+3^2+.....+8^2
= [n (n+1) (2n+1)]/6 (summation formula)
= 204
rectangles :
in a 2*2 chessboard, there are 9 rectangles (4 1*1s,1 2*2,2 2*1s, 2 1*2s)
for an n*n chessboard, there are 1^3 +2^3+3^3+.....+n^3 rectangles.
so for an 8*8 chessboard, there are, rectangles = 1^3+2^3+....+8^3
=> no. of rectangles = [{n^2}{(n+1)^2}/4
n=8, we get no. of rectangles = 1296
**No. of rectangles that are not squares in an 8*8 chessboard
=> 1296- 204
= 1092
in a chessboard,there are 8*8 squares.
in a 2*2 chessboard, there are 5 squares (4 small aquares, 1 big square).
similarly in an n*n chessboard, there are 1^2 +2^2+....+n^2 squares.
so in a 8*8 chessboard, n=8
=> no.of squares = 1^2 +2^2+3^2+.....+8^2
= [n (n+1) (2n+1)]/6 (summation formula)
= 204
rectangles :
in a 2*2 chessboard, there are 9 rectangles (4 1*1s,1 2*2,2 2*1s, 2 1*2s)
for an n*n chessboard, there are 1^3 +2^3+3^3+.....+n^3 rectangles.
so for an 8*8 chessboard, there are, rectangles = 1^3+2^3+....+8^3
=> no. of rectangles = [{n^2}{(n+1)^2}/4
n=8, we get no. of rectangles = 1296
**No. of rectangles that are not squares in an 8*8 chessboard
=> 1296- 204
= 1092
______________________
4.HOW to find the last non zero digit in x!
Last non-zero digit of 10! = 8
For 20! = 8*8 = 4 [6 is dropped]
This can continue for any number of 10s.
For example,
Last digit of 70! will be given by 8^7
For last digit of 8^7 = 2^21
=( 2^10)^2 * 2
= 76*2
=2
Therefore, last digit of 70! is 2.
If you are asked for 73!, then just multiply the above 2 with 71*72*73
or simply 2 with 1*2*3
Therefore, last non zero digit for 73! become 2*6 = 2 [ignoring 10s digit]
For 20! = 8*8 = 4 [6 is dropped]
This can continue for any number of 10s.
For example,
Last digit of 70! will be given by 8^7
For last digit of 8^7 = 2^21
=( 2^10)^2 * 2
= 76*2
=2
Therefore, last digit of 70! is 2.
If you are asked for 73!, then just multiply the above 2 with 71*72*73
or simply 2 with 1*2*3
Therefore, last non zero digit for 73! become 2*6 = 2 [ignoring 10s digit]
______________________
5. Pythagorus Triplets :-
How we can find different pythagorus triplets.
The Formula is n+[n/(2n+1)]
hence N can take values from 1,2.....
hence the first mixed fraction becomes 1[1/3] = 4/3
One side of the pythagorus theorem is denominator which is 3 and other side is numerator which is 4, the hypotenuse is 4+1 = 5
Hence the triplet is 3,4,5.
Lets take n = 3
2n+1 = 7
Mixed fraction becomes = 24/7
the first side becomes 7, the other side is 24 and hypotenuse is 24+1 = 25
hence 7,24,25.
all the other remaining pythagorus triplet is multiples of the triplets given by the above formula.
______________________________________________________________
6. Basic Formulae :-
(a) In a plane if there are n points of which no three are collinear, then
(b) In a plane if there are n points out of which m points are collinear, then
(c) The number of diagonals of a n sided polygon are nC2 – n = n × (n – 3)/2.
(d) The number of triangles that can be formed by joining the vertices of a n-sided polygon which has,
- The number of straight lines that can be formed by joining them isnC2.
- The number of triangles that can be formed by joining them is nC3.
- The number of polygons with k sides that can be formed by joining them isnCk.
(b) In a plane if there are n points out of which m points are collinear, then
- The number of straight lines that can be formed by joining them isnC2 – mC2 + 1.
- The number of triangles that can be formed by joining them is nC3 – mC3.
- The number of polygons with k sides that can be formed by joining them isnCk – mCk.
(c) The number of diagonals of a n sided polygon are nC2 – n = n × (n – 3)/2.
(d) The number of triangles that can be formed by joining the vertices of a n-sided polygon which has,
- Exactly one side common with that of the polygon are n × (n – 4).
- Exactly two sides common with that of the polygon are n.
- No side common with that of the polygon are n × (n – 4) × (n – 5)/6.
_____________________________________________________________________________________
7. Cauchy-Schwartz Equation:
If a , b , c , d are four real numbers, they always satisfy the relationship
(a^2+b^2)(c^2+d^2)>=(ac+bd)^2
This can be generalized to a large number of variables as
(a1^2+a2^2+a3^2+.....)(b1^2+b2^2+b3^2+....)>=(a1b1 +a2b2+a3b3+....)^2
Questions:
1>Find the least value of X^2+Y^2+z^2 if X+2y+3Z=14
Sol:-->(X^2+Y^2+z^2)(1^2+2^2+3^2)>=(X*1+Y*2+z*3)^2
hence , min value= 14^2/14=14
(a^2+b^2)(c^2+d^2)>=(ac+bd)^2
This can be generalized to a large number of variables as
(a1^2+a2^2+a3^2+.....)(b1^2+b2^2+b3^2+....)>=(a1b1 +a2b2+a3b3+....)^2
Questions:
1>Find the least value of X^2+Y^2+z^2 if X+2y+3Z=14
Sol:-->(X^2+Y^2+z^2)(1^2+2^2+3^2)>=(X*1+Y*2+z*3)^2
hence , min value= 14^2/14=14
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8. Basic Formula
The number of regions into which n straight lines divide the plane = n(n+1)/2 +1.
The number of unbounded regions = 2n.
The number of unbounded regions = 2n.
if a circle is drawn and n straight lines are drawn, then all the regions are bounded which is n(n+1)/2 + 1
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