Introduction to generalized permutations and combinations:
we have 2 formulas ncr and npr
If there are 'n' things, out of which 'r' things are to be selected, the number of ways in which 'r' things out of 'n' things are selected is denoted by nCr. It is called Combination.
But if the selected 'r' things are arranged in a definite order, then we call it npr. It is called Permutation.
Generalized permutation and combinations are done using the nCr and nPr formulas
Generalized Permutations and Combinations Problems for n things not all different has the following formula.
Out of 'n' things, if 'p' things are of one kind, 'q' things are of another kind and 'r' things are of a third kind, then
the generalized formula for this permutation is N .p!.q!.r! = n! where N = number of permutations
The generalized formula for this permutation becomes N = n!
-------
p!q!r!
Generalized Permutations and Combinations-permutation:-
The number of permutation of n things taken r at a time =npr= n(n-1)(n-2) .............. (n-r+1)
If r = n , then npr = npn
Then we get npn= n(n-1) (n-2) ............ 3.2.1
This continued product is denoted by a factorial n! Therefore npn= n! ! sign is called factorial
T he formula for npr = `|__` n L is also called factorial
------
`|__` n-r
Let us do a couple of problems on permutations.
1) Find 10p4
Answer 10p4= `|__` 10 `|__` 10 10x9x8x7x6x5x4x3x2x1 10x9x8x7 = 5040
------- = ------------ = ------------------------------- =
`|__` 10-4 `|__` 6 6x5x4x3x2x1
2) How many 6 digit numbers can be formed using the digits 4,5,6,7,8,8, no digit being repeated in each number
Answer the permutatio is 6p6 = 6! = 6x5x4x3x2x1 = 720 numbers can be formed
3) In how many ways can 3 white balls, 4 red balls and 5 blue balls can be arranged in a row so as to keep all the
balls of the same color together?
Answer:-Now we consider the 4 red balls as a unit because they must be together. Similarly consider the 5 blue balls
as a unit and the 3 white balls as a unit.
Now we get 3 units which can be arranged in 3p3 ways = 3x2x1= 6 ways.
Next 4 red balls can be arranged among themselves in 4p4 ways = 4! = 4x3x2x1= 24 ways
Next 5 blue balls can be arranged in 5p5 ways = 5! = 5x4x3x2x1 = 120 ways
Next 3 white balls can be arranged in 3p3 = 6 ways
The require no of arrangements = 6 x24x 120 x6 = 103,680 ways
Generalized Permutations and Combinations-combinations:-
The formula for nCr = n!
-------
(n-r)! r!
It can also be written as nCr = nPr
-------
r!
Let us do a few problems on combinations.
Problem 4 Find 10C4
10C4 = 10! 10x9x8x7x6x5x4x3x2x1 10x9x8x7
--------- --------------------------------- = ------------- = 210
(10-4)! 4! [ 6x5x4x3x2x1][ 4x3x2x1] 4x3x2x1
Problem 5 A committee of 3 teachers and 2 students is to be formed from 5 teachers and 10 students. In how many ways
can this be done?
Teachers n= 5 and r = 3 so nCr = 5C3 = 5x4x3
------- = 10 combinations
1x2x3
Students n=10 r = 2 so nCr = 10C2= 10 x9
------- = 45 combinations
1x2
Total combinations = 10 x 45 = 450 combinations
Circular Permutation
Circular permutation is an arrangement in which the things are arranged in a circle.
let us arrange a,b,c,d in a circular arrangement.
We can arrange them in 4! ways = 24 ways
But when we arrange them in a circle we get 4 circular arrangements . Hence 24 ÷ 4 = 6 = 3.2.1 = 3!
Thus the generalized permutation formula for circular permutation is n!
-- = (n-1)!
n
This circular permutation formula gets modified in case of beads and necklaces. since there is no difference
in clockwise or anticlockwise arrangement in the case of beads or necklaces of circular permutation, the
formula becomes nPr = (n-1)!
-------
2
Find the number of ways in which 5 beads can be strund in a ring
This is circular permutation of beads. Hence the formula is
Number of permutations = (5-1)! 4! 24
------- = ---- = --- = 12
2 2 2
But when we do a problem like find the number of ways in which 5 people can be arranged at a round table, I have recently faced lot of problem while learning how to graph an equation, But thank to online resources of math which helped me to learn myself easily on net.
we must remember that the 5 people are not alike then the number of permutation= (5-1)! = 4! = 24
Generalized Permutation and Combination Problems:-
Let us give the student a few exercise problems
1. Find 7C2 and 7C5 What do you infer from this? Answer : 7C2 = 21 7C5 = 21Both are equal
2. Find 8C3 Answer : 56
3. Find the number of diagonals in a decagon Answer ; 35
4. Find the number of triangles formed by the vertices of a hexagon Answer : 20
5. A man has 3 friends. In how many ways can he invite them to a dinnerAnswer : 1
6. Find 5P3 Answer : 60
7. If nP2 = 30 find n Answer: n=6
8. In how many ways can 6 different beads can be strung into a necklace Answer: 60
we have 2 formulas ncr and npr
If there are 'n' things, out of which 'r' things are to be selected, the number of ways in which 'r' things out of 'n' things are selected is denoted by nCr. It is called Combination.
But if the selected 'r' things are arranged in a definite order, then we call it npr. It is called Permutation.
Generalized permutation and combinations are done using the nCr and nPr formulas
Generalized Permutations and Combinations Problems for n things not all different has the following formula.
Out of 'n' things, if 'p' things are of one kind, 'q' things are of another kind and 'r' things are of a third kind, then
the generalized formula for this permutation is N .p!.q!.r! = n! where N = number of permutations
The generalized formula for this permutation becomes N = n!
-------
p!q!r!
Generalized Permutations and Combinations-permutation:-
The number of permutation of n things taken r at a time =npr= n(n-1)(n-2) .............. (n-r+1)
If r = n , then npr = npn
Then we get npn= n(n-1) (n-2) ............ 3.2.1
This continued product is denoted by a factorial n! Therefore npn= n! ! sign is called factorial
T he formula for npr = `|__` n L is also called factorial
------
`|__` n-r
Let us do a couple of problems on permutations.
1) Find 10p4
Answer 10p4= `|__` 10 `|__` 10 10x9x8x7x6x5x4x3x2x1 10x9x8x7 = 5040
------- = ------------ = ------------------------------- =
`|__` 10-4 `|__` 6 6x5x4x3x2x1
2) How many 6 digit numbers can be formed using the digits 4,5,6,7,8,8, no digit being repeated in each number
Answer the permutatio is 6p6 = 6! = 6x5x4x3x2x1 = 720 numbers can be formed
3) In how many ways can 3 white balls, 4 red balls and 5 blue balls can be arranged in a row so as to keep all the
balls of the same color together?
Answer:-Now we consider the 4 red balls as a unit because they must be together. Similarly consider the 5 blue balls
as a unit and the 3 white balls as a unit.
Now we get 3 units which can be arranged in 3p3 ways = 3x2x1= 6 ways.
Next 4 red balls can be arranged among themselves in 4p4 ways = 4! = 4x3x2x1= 24 ways
Next 5 blue balls can be arranged in 5p5 ways = 5! = 5x4x3x2x1 = 120 ways
Next 3 white balls can be arranged in 3p3 = 6 ways
The require no of arrangements = 6 x24x 120 x6 = 103,680 ways
Generalized Permutations and Combinations-combinations:-
The formula for nCr = n!
-------
(n-r)! r!
It can also be written as nCr = nPr
-------
r!
Let us do a few problems on combinations.
Problem 4 Find 10C4
10C4 = 10! 10x9x8x7x6x5x4x3x2x1 10x9x8x7
--------- --------------------------------- = ------------- = 210
(10-4)! 4! [ 6x5x4x3x2x1][ 4x3x2x1] 4x3x2x1
Problem 5 A committee of 3 teachers and 2 students is to be formed from 5 teachers and 10 students. In how many ways
can this be done?
Teachers n= 5 and r = 3 so nCr = 5C3 = 5x4x3
------- = 10 combinations
1x2x3
Students n=10 r = 2 so nCr = 10C2= 10 x9
------- = 45 combinations
1x2
Total combinations = 10 x 45 = 450 combinations
Circular Permutation
Circular permutation is an arrangement in which the things are arranged in a circle.
let us arrange a,b,c,d in a circular arrangement.
We can arrange them in 4! ways = 24 ways
But when we arrange them in a circle we get 4 circular arrangements . Hence 24 ÷ 4 = 6 = 3.2.1 = 3!
Thus the generalized permutation formula for circular permutation is n!
-- = (n-1)!
n
This circular permutation formula gets modified in case of beads and necklaces. since there is no difference
in clockwise or anticlockwise arrangement in the case of beads or necklaces of circular permutation, the
formula becomes nPr = (n-1)!
-------
2
Find the number of ways in which 5 beads can be strund in a ring
This is circular permutation of beads. Hence the formula is
Number of permutations = (5-1)! 4! 24
------- = ---- = --- = 12
2 2 2
But when we do a problem like find the number of ways in which 5 people can be arranged at a round table, I have recently faced lot of problem while learning how to graph an equation, But thank to online resources of math which helped me to learn myself easily on net.
we must remember that the 5 people are not alike then the number of permutation= (5-1)! = 4! = 24
Generalized Permutation and Combination Problems:-
Let us give the student a few exercise problems
1. Find 7C2 and 7C5 What do you infer from this? Answer : 7C2 = 21 7C5 = 21Both are equal
2. Find 8C3 Answer : 56
3. Find the number of diagonals in a decagon Answer ; 35
4. Find the number of triangles formed by the vertices of a hexagon Answer : 20
5. A man has 3 friends. In how many ways can he invite them to a dinnerAnswer : 1
6. Find 5P3 Answer : 60
7. If nP2 = 30 find n Answer: n=6
8. In how many ways can 6 different beads can be strung into a necklace Answer: 60
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