Detailed Notes: Permutations and Combinations | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


PERMUTATION AND 
COMBINATION 
The most fundamental application of mathematics is counting. There are many natural 
methods used for counting 
This chapter is dealing with various known techniques those are much faster than the 
usual counting methods 
We mainly focus, our methods, on counting the number of arrangements (Permutations) 
and the number of selections (combinations), even although we may use these 
techniques for counting in some other situations also 
Let us start with a simple problem 
A group ?? 1
 of 3 circles ?? 1
, ?? 2
, ?? 3
 having different centers are situated in such a way that ?? 2
 
lie entirely inside C
1
; C
3
 lie entirely inside C
2
. Another group G
2
 of 4 circles 
C
1
 
'
, C
2
 
'
, C
3
 
'
, C
4
 
'
 are also situated in a similar fashion. The two groups of circles are in 
such a way that each member of ?? 1
 intersect with every member of G
2
, as shown in the 
following figure 
 
(i) How many centres the circles altogether has ? 
(ii) How many common chords are obtained? 
The answer to the first part is " 3 + 4 = 7 " and answer to the second part is "3 × 4 = 12 
". The method in which we calculated first part of the problem is called as "addition rule" 
and the method we used to calculate its second part is called as the "multiplication rule". 
These rules altogether are the most important tools in counting, popularly known as "the 
fundamental counting principle". 
 
Page 2


PERMUTATION AND 
COMBINATION 
The most fundamental application of mathematics is counting. There are many natural 
methods used for counting 
This chapter is dealing with various known techniques those are much faster than the 
usual counting methods 
We mainly focus, our methods, on counting the number of arrangements (Permutations) 
and the number of selections (combinations), even although we may use these 
techniques for counting in some other situations also 
Let us start with a simple problem 
A group ?? 1
 of 3 circles ?? 1
, ?? 2
, ?? 3
 having different centers are situated in such a way that ?? 2
 
lie entirely inside C
1
; C
3
 lie entirely inside C
2
. Another group G
2
 of 4 circles 
C
1
 
'
, C
2
 
'
, C
3
 
'
, C
4
 
'
 are also situated in a similar fashion. The two groups of circles are in 
such a way that each member of ?? 1
 intersect with every member of G
2
, as shown in the 
following figure 
 
(i) How many centres the circles altogether has ? 
(ii) How many common chords are obtained? 
The answer to the first part is " 3 + 4 = 7 " and answer to the second part is "3 × 4 = 12 
". The method in which we calculated first part of the problem is called as "addition rule" 
and the method we used to calculate its second part is called as the "multiplication rule". 
These rules altogether are the most important tools in counting, popularly known as "the 
fundamental counting principle". 
 
Fundamental counting principle : 
Suppose that an operation O
1
 can be done in m different ways and another operation O
2
 
can be done in n different ways. 
(i) Addition rule : The number of ways in which we can do exactly one of the operations 
O
1
, O
2
 is m + n 
(ii) Multiplication rule : The number of ways in which we can do both the operations 
O
1
, O
2
 is mn. 
Note : The addition rule is true only when O
1
&O
2
 are mutually exclusive and 
multiplication rule is true only when O
1
&O
2
 are independent (The reader will understand 
the concepts of mutual exclusiveness and independence, in the due course) 
Problem 1: There are 8 buses running from Kota to Jaipur and 10 buses running from 
Jaipur to Delhi. In how many ways a person can travel from Kota to Delhi via Jaipur by 
bus? 
Solution : Let ?? 1
 be the event of travelling from Kota to Jaipur &?? 2
 be the event of 
travelling from Jaipur to 
Delhi by the person. 
E
1
 can happen in 8 ways and E
2
 can happen in 10 ways 
Since both the events ?? 1
 and ?? 2
 are to be happened in order, simultaneously, the number 
of ways = 8 × 10 = 80. 
Problem 2 : How many numbers between 10 and 10,000 can be formed by using the 
digits 1, 2, 3, 4, 5 if 
(i) No digit is repeated in any number. 
(ii) Digits can be repeated. 
Solution : (i) Number of two digit numbers = 5 × 4 = 20 
Number of three digit numbers = 5 × 4 × 3 = 60 
Number of four digit numbers = 5 × 4 × 3 × 2 = 120 
Total = 200 
(ii) Number of two digit numbers = 5 × 5 = 25 
Number of three digit numbers = 5 × 5 × 5 = 125 
Number of four digit numbers = 5 × 5 × 5 × 5 = 625 
Total = 775 
Page 3


PERMUTATION AND 
COMBINATION 
The most fundamental application of mathematics is counting. There are many natural 
methods used for counting 
This chapter is dealing with various known techniques those are much faster than the 
usual counting methods 
We mainly focus, our methods, on counting the number of arrangements (Permutations) 
and the number of selections (combinations), even although we may use these 
techniques for counting in some other situations also 
Let us start with a simple problem 
A group ?? 1
 of 3 circles ?? 1
, ?? 2
, ?? 3
 having different centers are situated in such a way that ?? 2
 
lie entirely inside C
1
; C
3
 lie entirely inside C
2
. Another group G
2
 of 4 circles 
C
1
 
'
, C
2
 
'
, C
3
 
'
, C
4
 
'
 are also situated in a similar fashion. The two groups of circles are in 
such a way that each member of ?? 1
 intersect with every member of G
2
, as shown in the 
following figure 
 
(i) How many centres the circles altogether has ? 
(ii) How many common chords are obtained? 
The answer to the first part is " 3 + 4 = 7 " and answer to the second part is "3 × 4 = 12 
". The method in which we calculated first part of the problem is called as "addition rule" 
and the method we used to calculate its second part is called as the "multiplication rule". 
These rules altogether are the most important tools in counting, popularly known as "the 
fundamental counting principle". 
 
Fundamental counting principle : 
Suppose that an operation O
1
 can be done in m different ways and another operation O
2
 
can be done in n different ways. 
(i) Addition rule : The number of ways in which we can do exactly one of the operations 
O
1
, O
2
 is m + n 
(ii) Multiplication rule : The number of ways in which we can do both the operations 
O
1
, O
2
 is mn. 
Note : The addition rule is true only when O
1
&O
2
 are mutually exclusive and 
multiplication rule is true only when O
1
&O
2
 are independent (The reader will understand 
the concepts of mutual exclusiveness and independence, in the due course) 
Problem 1: There are 8 buses running from Kota to Jaipur and 10 buses running from 
Jaipur to Delhi. In how many ways a person can travel from Kota to Delhi via Jaipur by 
bus? 
Solution : Let ?? 1
 be the event of travelling from Kota to Jaipur &?? 2
 be the event of 
travelling from Jaipur to 
Delhi by the person. 
E
1
 can happen in 8 ways and E
2
 can happen in 10 ways 
Since both the events ?? 1
 and ?? 2
 are to be happened in order, simultaneously, the number 
of ways = 8 × 10 = 80. 
Problem 2 : How many numbers between 10 and 10,000 can be formed by using the 
digits 1, 2, 3, 4, 5 if 
(i) No digit is repeated in any number. 
(ii) Digits can be repeated. 
Solution : (i) Number of two digit numbers = 5 × 4 = 20 
Number of three digit numbers = 5 × 4 × 3 = 60 
Number of four digit numbers = 5 × 4 × 3 × 2 = 120 
Total = 200 
(ii) Number of two digit numbers = 5 × 5 = 25 
Number of three digit numbers = 5 × 5 × 5 = 125 
Number of four digit numbers = 5 × 5 × 5 × 5 = 625 
Total = 775 
Arrangements : 
If n???? denotes the number of permutations (arrangements) of ?? different things, taking ?? 
at a time, then 
PPr
?? = ?? (?? - 1)(?? - 2)… . . (?? - ?? + 1) =
?? !
(?? - ?? )!
 
NOTE : (i) Factorials of negative integers are not defined. 
(ii) 0! = 1! = 1 
(iii)  
?? ?? ?? = ?? ! = ?? · (?? - 1)! 
(iv) (2?? )! = 2
?? · ?? ! [1 · 3 · 5.7 … (2?? - 1)] 
Problem 3 : How many three digit can be formed using the digits 1, 2, 3, 4, 5, without 
repetition of digits? How many of these are even? 
Solution : Three places are to be filled with 5 different objects. 
?  Number of ways = 
5
?? 3
= 5 × 4 × 3 = 60 
For the 2nd part, unit digit can be filled in two ways & the remaining two digits can be 
filled in  
4
P
2
 ways. 
?  Number of even numbers = 2 × 
4
P
2
= 24. 
Problem 4: If all the letters of the word 'QUEST' are arranged in all possible ways and 
put in dictionary order, then find the rank of the given word. 
Solution :   Number of words beginning with E = 
4
P
4
= 24 
Number of words beginning with QE = 
3
P
3
= 6 
Number of words beginning with QS = 6 
Number of words beginning withQT = 6. 
Next word is 'QUEST' 
?  its rank is 24 + 6 + 6 + 6 + 1 = 43. 
Combination : 
If  
?? ?? ?? denotes the number of combinations (selections) of ?? different things taken ?? at a 
time, then  
?? ?? ?? =
?? !
?? !(?? -?? )!
=
 
?? ?? ?? ?? !
 where ?? = ?? ; ?? ? ?? and ?? ? ?? . 
NOTE : (i)  
?? ?? ?? = 
?? ?? ?? -?? 
Page 4


PERMUTATION AND 
COMBINATION 
The most fundamental application of mathematics is counting. There are many natural 
methods used for counting 
This chapter is dealing with various known techniques those are much faster than the 
usual counting methods 
We mainly focus, our methods, on counting the number of arrangements (Permutations) 
and the number of selections (combinations), even although we may use these 
techniques for counting in some other situations also 
Let us start with a simple problem 
A group ?? 1
 of 3 circles ?? 1
, ?? 2
, ?? 3
 having different centers are situated in such a way that ?? 2
 
lie entirely inside C
1
; C
3
 lie entirely inside C
2
. Another group G
2
 of 4 circles 
C
1
 
'
, C
2
 
'
, C
3
 
'
, C
4
 
'
 are also situated in a similar fashion. The two groups of circles are in 
such a way that each member of ?? 1
 intersect with every member of G
2
, as shown in the 
following figure 
 
(i) How many centres the circles altogether has ? 
(ii) How many common chords are obtained? 
The answer to the first part is " 3 + 4 = 7 " and answer to the second part is "3 × 4 = 12 
". The method in which we calculated first part of the problem is called as "addition rule" 
and the method we used to calculate its second part is called as the "multiplication rule". 
These rules altogether are the most important tools in counting, popularly known as "the 
fundamental counting principle". 
 
Fundamental counting principle : 
Suppose that an operation O
1
 can be done in m different ways and another operation O
2
 
can be done in n different ways. 
(i) Addition rule : The number of ways in which we can do exactly one of the operations 
O
1
, O
2
 is m + n 
(ii) Multiplication rule : The number of ways in which we can do both the operations 
O
1
, O
2
 is mn. 
Note : The addition rule is true only when O
1
&O
2
 are mutually exclusive and 
multiplication rule is true only when O
1
&O
2
 are independent (The reader will understand 
the concepts of mutual exclusiveness and independence, in the due course) 
Problem 1: There are 8 buses running from Kota to Jaipur and 10 buses running from 
Jaipur to Delhi. In how many ways a person can travel from Kota to Delhi via Jaipur by 
bus? 
Solution : Let ?? 1
 be the event of travelling from Kota to Jaipur &?? 2
 be the event of 
travelling from Jaipur to 
Delhi by the person. 
E
1
 can happen in 8 ways and E
2
 can happen in 10 ways 
Since both the events ?? 1
 and ?? 2
 are to be happened in order, simultaneously, the number 
of ways = 8 × 10 = 80. 
Problem 2 : How many numbers between 10 and 10,000 can be formed by using the 
digits 1, 2, 3, 4, 5 if 
(i) No digit is repeated in any number. 
(ii) Digits can be repeated. 
Solution : (i) Number of two digit numbers = 5 × 4 = 20 
Number of three digit numbers = 5 × 4 × 3 = 60 
Number of four digit numbers = 5 × 4 × 3 × 2 = 120 
Total = 200 
(ii) Number of two digit numbers = 5 × 5 = 25 
Number of three digit numbers = 5 × 5 × 5 = 125 
Number of four digit numbers = 5 × 5 × 5 × 5 = 625 
Total = 775 
Arrangements : 
If n???? denotes the number of permutations (arrangements) of ?? different things, taking ?? 
at a time, then 
PPr
?? = ?? (?? - 1)(?? - 2)… . . (?? - ?? + 1) =
?? !
(?? - ?? )!
 
NOTE : (i) Factorials of negative integers are not defined. 
(ii) 0! = 1! = 1 
(iii)  
?? ?? ?? = ?? ! = ?? · (?? - 1)! 
(iv) (2?? )! = 2
?? · ?? ! [1 · 3 · 5.7 … (2?? - 1)] 
Problem 3 : How many three digit can be formed using the digits 1, 2, 3, 4, 5, without 
repetition of digits? How many of these are even? 
Solution : Three places are to be filled with 5 different objects. 
?  Number of ways = 
5
?? 3
= 5 × 4 × 3 = 60 
For the 2nd part, unit digit can be filled in two ways & the remaining two digits can be 
filled in  
4
P
2
 ways. 
?  Number of even numbers = 2 × 
4
P
2
= 24. 
Problem 4: If all the letters of the word 'QUEST' are arranged in all possible ways and 
put in dictionary order, then find the rank of the given word. 
Solution :   Number of words beginning with E = 
4
P
4
= 24 
Number of words beginning with QE = 
3
P
3
= 6 
Number of words beginning with QS = 6 
Number of words beginning withQT = 6. 
Next word is 'QUEST' 
?  its rank is 24 + 6 + 6 + 6 + 1 = 43. 
Combination : 
If  
?? ?? ?? denotes the number of combinations (selections) of ?? different things taken ?? at a 
time, then  
?? ?? ?? =
?? !
?? !(?? -?? )!
=
 
?? ?? ?? ?? !
 where ?? = ?? ; ?? ? ?? and ?? ? ?? . 
NOTE : (i)  
?? ?? ?? = 
?? ?? ?? -?? 
(ii)  
?? ?? ?? + 
?? ?? ?? -1
= 
?? +1
?? ?? 
(iii)  
?? ?? ?? = 0 if ?? ? {0,1,2,3 … … . , ?? } 
Problem 7 : There are fifteen players for a cricket match. 
(i) In how many ways the 11 players can be selected? 
(ii) In how many ways the 11 players can be selected including a particular player? 
(iii) In how many ways the 11 players can be selected excluding two particular players? 
Solution: (i) 11 players are to be selected from 15 
Number of ways = 
15
C
11
= 1365. 
(ii) Since one player is already included, we have to select 10 from the remaining 14 
Number of ways = 
14
C
10
= 1001. 
(iii) Since two players are to be excluded, we have to select 11 from the remaining 13 . 
Number of ways = 
13
C
11
= 78. 
Problem 8 : If  
49
C
3?? -2
= 
49
C
2?? +1
, find ' ?? '. 
Solution:   
?? ?? ?? = 
?? ?? ?? if either ?? = ?? or ?? + ?? = ?? . 
 Thus 3?? - 2 = 2?? + 1
 or 3?? - 2 + 2?? + 1 = 49
? ?? = 3,10
 ? 
?? = 3
5?? - 1 = 49
 ? ?? = 10 
Problem 9: A regular polygon has 20 sides. How many triangles can be drawn by using 
the vertices, but not using the sides? 
Solution: The first vertex can be selected in 20 ways. The remaining two are to be 
selected from 17 vertices so that they are not consecutive. This can be done in  
17
C
2
- 16 
ways. 
?  The total number of ways = 20 × ( 
17
C
2
- 16) 
But in this method, each selection is repeated thrice. 
?  Number of triangles =
20 × ( 
17
C
2
- 16)
3
= 800.  
Problem 10: 15 persons are sitting in a row. In how many ways we can select three of 
them if adjacent persons are not selected ? 
Solution : Let ?? 1
, ?? 2
, ?? 3
, ?? 4
, ?? 5
, ?? 6
, ?? 7
, ?? 8
, ?? 9
, ?? 10
, ?? 11
, ?? 12
, ?? 13
, ?? 14
, ?? 15
 be the persons sitting in 
this order. If three are selected (non consecutive) then 12 are left out. 
Page 5


PERMUTATION AND 
COMBINATION 
The most fundamental application of mathematics is counting. There are many natural 
methods used for counting 
This chapter is dealing with various known techniques those are much faster than the 
usual counting methods 
We mainly focus, our methods, on counting the number of arrangements (Permutations) 
and the number of selections (combinations), even although we may use these 
techniques for counting in some other situations also 
Let us start with a simple problem 
A group ?? 1
 of 3 circles ?? 1
, ?? 2
, ?? 3
 having different centers are situated in such a way that ?? 2
 
lie entirely inside C
1
; C
3
 lie entirely inside C
2
. Another group G
2
 of 4 circles 
C
1
 
'
, C
2
 
'
, C
3
 
'
, C
4
 
'
 are also situated in a similar fashion. The two groups of circles are in 
such a way that each member of ?? 1
 intersect with every member of G
2
, as shown in the 
following figure 
 
(i) How many centres the circles altogether has ? 
(ii) How many common chords are obtained? 
The answer to the first part is " 3 + 4 = 7 " and answer to the second part is "3 × 4 = 12 
". The method in which we calculated first part of the problem is called as "addition rule" 
and the method we used to calculate its second part is called as the "multiplication rule". 
These rules altogether are the most important tools in counting, popularly known as "the 
fundamental counting principle". 
 
Fundamental counting principle : 
Suppose that an operation O
1
 can be done in m different ways and another operation O
2
 
can be done in n different ways. 
(i) Addition rule : The number of ways in which we can do exactly one of the operations 
O
1
, O
2
 is m + n 
(ii) Multiplication rule : The number of ways in which we can do both the operations 
O
1
, O
2
 is mn. 
Note : The addition rule is true only when O
1
&O
2
 are mutually exclusive and 
multiplication rule is true only when O
1
&O
2
 are independent (The reader will understand 
the concepts of mutual exclusiveness and independence, in the due course) 
Problem 1: There are 8 buses running from Kota to Jaipur and 10 buses running from 
Jaipur to Delhi. In how many ways a person can travel from Kota to Delhi via Jaipur by 
bus? 
Solution : Let ?? 1
 be the event of travelling from Kota to Jaipur &?? 2
 be the event of 
travelling from Jaipur to 
Delhi by the person. 
E
1
 can happen in 8 ways and E
2
 can happen in 10 ways 
Since both the events ?? 1
 and ?? 2
 are to be happened in order, simultaneously, the number 
of ways = 8 × 10 = 80. 
Problem 2 : How many numbers between 10 and 10,000 can be formed by using the 
digits 1, 2, 3, 4, 5 if 
(i) No digit is repeated in any number. 
(ii) Digits can be repeated. 
Solution : (i) Number of two digit numbers = 5 × 4 = 20 
Number of three digit numbers = 5 × 4 × 3 = 60 
Number of four digit numbers = 5 × 4 × 3 × 2 = 120 
Total = 200 
(ii) Number of two digit numbers = 5 × 5 = 25 
Number of three digit numbers = 5 × 5 × 5 = 125 
Number of four digit numbers = 5 × 5 × 5 × 5 = 625 
Total = 775 
Arrangements : 
If n???? denotes the number of permutations (arrangements) of ?? different things, taking ?? 
at a time, then 
PPr
?? = ?? (?? - 1)(?? - 2)… . . (?? - ?? + 1) =
?? !
(?? - ?? )!
 
NOTE : (i) Factorials of negative integers are not defined. 
(ii) 0! = 1! = 1 
(iii)  
?? ?? ?? = ?? ! = ?? · (?? - 1)! 
(iv) (2?? )! = 2
?? · ?? ! [1 · 3 · 5.7 … (2?? - 1)] 
Problem 3 : How many three digit can be formed using the digits 1, 2, 3, 4, 5, without 
repetition of digits? How many of these are even? 
Solution : Three places are to be filled with 5 different objects. 
?  Number of ways = 
5
?? 3
= 5 × 4 × 3 = 60 
For the 2nd part, unit digit can be filled in two ways & the remaining two digits can be 
filled in  
4
P
2
 ways. 
?  Number of even numbers = 2 × 
4
P
2
= 24. 
Problem 4: If all the letters of the word 'QUEST' are arranged in all possible ways and 
put in dictionary order, then find the rank of the given word. 
Solution :   Number of words beginning with E = 
4
P
4
= 24 
Number of words beginning with QE = 
3
P
3
= 6 
Number of words beginning with QS = 6 
Number of words beginning withQT = 6. 
Next word is 'QUEST' 
?  its rank is 24 + 6 + 6 + 6 + 1 = 43. 
Combination : 
If  
?? ?? ?? denotes the number of combinations (selections) of ?? different things taken ?? at a 
time, then  
?? ?? ?? =
?? !
?? !(?? -?? )!
=
 
?? ?? ?? ?? !
 where ?? = ?? ; ?? ? ?? and ?? ? ?? . 
NOTE : (i)  
?? ?? ?? = 
?? ?? ?? -?? 
(ii)  
?? ?? ?? + 
?? ?? ?? -1
= 
?? +1
?? ?? 
(iii)  
?? ?? ?? = 0 if ?? ? {0,1,2,3 … … . , ?? } 
Problem 7 : There are fifteen players for a cricket match. 
(i) In how many ways the 11 players can be selected? 
(ii) In how many ways the 11 players can be selected including a particular player? 
(iii) In how many ways the 11 players can be selected excluding two particular players? 
Solution: (i) 11 players are to be selected from 15 
Number of ways = 
15
C
11
= 1365. 
(ii) Since one player is already included, we have to select 10 from the remaining 14 
Number of ways = 
14
C
10
= 1001. 
(iii) Since two players are to be excluded, we have to select 11 from the remaining 13 . 
Number of ways = 
13
C
11
= 78. 
Problem 8 : If  
49
C
3?? -2
= 
49
C
2?? +1
, find ' ?? '. 
Solution:   
?? ?? ?? = 
?? ?? ?? if either ?? = ?? or ?? + ?? = ?? . 
 Thus 3?? - 2 = 2?? + 1
 or 3?? - 2 + 2?? + 1 = 49
? ?? = 3,10
 ? 
?? = 3
5?? - 1 = 49
 ? ?? = 10 
Problem 9: A regular polygon has 20 sides. How many triangles can be drawn by using 
the vertices, but not using the sides? 
Solution: The first vertex can be selected in 20 ways. The remaining two are to be 
selected from 17 vertices so that they are not consecutive. This can be done in  
17
C
2
- 16 
ways. 
?  The total number of ways = 20 × ( 
17
C
2
- 16) 
But in this method, each selection is repeated thrice. 
?  Number of triangles =
20 × ( 
17
C
2
- 16)
3
= 800.  
Problem 10: 15 persons are sitting in a row. In how many ways we can select three of 
them if adjacent persons are not selected ? 
Solution : Let ?? 1
, ?? 2
, ?? 3
, ?? 4
, ?? 5
, ?? 6
, ?? 7
, ?? 8
, ?? 9
, ?? 10
, ?? 11
, ?? 12
, ?? 13
, ?? 14
, ?? 15
 be the persons sitting in 
this order. If three are selected (non consecutive) then 12 are left out. 
Let ?? , ?? , ?? , ?? , ?? , ?? , ?? , ?? , ?? , ?? , ?? , ?? be the left out &?? , ?? , ?? be the selected. The number of 
ways in which these 3 q's can be placed into the 13 positions between the P's (including 
extremes) is the number ways of required selection. 
Thus number of ways = 
13
C
3
= 286. 
Problem 11: In how many ways we can select 4 letters from the letters of the word 
MISSISSIPPI? 
Solution: M 
II I I 
SSSS 
PP 
Number of ways of selecting 4 alike letters = 
2
C
1
= 2. 
Number of ways of selecting 3 alike and 1 different letters = 
2
C
1
× 
3
C
1
= 6 
Number of ways of selecting 2 alike and 2 alike letters = 
3
C
2
= 3 
Number of ways of selecting 2 alike & 2 different = 
3
C
1
× 
3
C
2
= 9 
Number of ways of selecting 4 different = 
4
C
4
= 1 
Total number of ways = 2 + 6 + 3 + 9 + 1 = 21 
Arrangement of ?? things, those are not 
all different : 
The number of permutations of ' ?? ' things, taken all at a time, when ' ?? ' of them are 
same & of one type, ?? of them are same & of second type, 'r' of them are same & of a third 
type & the remaining ?? - (?? + ?? + ?? ) things are all different, is 
?? !
?? !?? !?? !
. 
Problem 12 : In how many ways we can arrange 3 red flowers, 4 yellow flowers and 5 
white flowers in a row? In how many ways this is possible if the white flowers are to be 
separated in any arrangement? (Flowers of same colour are identical). 
Solution:   Total we have 12 flowers 3 red, 4 yellow and 5 white. 
Number of arrangements =
12!
3!4!5!
= 27720. 
For the second part, first arrange 3 red & 4 yellow 
This can be done in 
7!
3!4!
= 35 ways