PERMUTATIONS AND COMBINATIONS
MAIN CONCEPTS AND RESULTS
** Fundamental principle of counting, or( the multiplication principle): “If an event can occur in m
different ways, following which another event can occur in n different ways, then the total number of
occurrence of the events in the given order is m × n.”
**Factorial notation The notation n! represents the product of first n natural numbers, i.e., the product
1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this symbol as „n factorial‟.
Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !
n ! = n (n – 1) !
= n (n – 1) (n – 2) ! [provided (n ≥ 2)]
= n (n – 1) (n – 2) (n – 3) ! [provided (n ≥ 3)]
**Permutations A permutation is an arrangement in a definite order of a number of objects taken
some or all at a time.
** The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects
do not
repeat is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by
n!
P (n , r) OR n Pr ,0 r n
n r !
** n P0 1 n Pn
** The number of permutations of n different objects taken r at a time, where repetition is allowed, is
nr.
** The number of permutations of n objects, where p1 objects are of one kind, p2 are of second kind, ...,
n!
pk are of kth kind and the rest, if any, are of different kind is .
p1!p2!...pk !
** The number of permutations of an dissimilar things taken all at a time along a circle is ( n -1) !.
** The number of ways of arranging a distinct objects along a circle when clockwise and anticlockwise
1
arrangements are considered alike is (n -1) !.
2
** The number of ways in which (m + n) different things can be divided into two groups containing
( m n) !
m and n things is .
m ! n!
n!
Combination of n different objects taken r at a time, denoted by n C r .
r!n r !
** n Pr n Cr r! , 0 r n
* * n C0 1 n C n
n
C1 n n Cn 1
n n 1 n
n
C2 C2
2!
n n 1n 3 n
n
C3 Cn 3
3!
** n Cr n Cs r s or r s n
** n Cr n Cr 1 n 1Cr
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, II. Some illustrations/Examples
MCQs
Q.1 How many three digit numbers are there with all distinct digits1,2,3,4,5,6,7,8,9,0 ?
(a) 458 (b) 568 (c) 648 (d) 748
Sol. (c)9.9.8= 648
9 9 8
Q.2The H.C.F. of 6!, 8!, 9!, 11!. is
(a) 6! (b) 8! (c) 9! (d) 11!
Sol. (a) 6!
Q.3 A polygon has 14 sides then number of its diagonals are:
(a) 91 (b) 77 (c) 28 (d) none of these
Sol. (b) 77
Let the no. of sides is n.
Then no. of its diagonals is = 14C2 – 14= 91 –14 = 77
Q.4 IfnC5 = nC7, find n.
(a) 12 (b) 15 (c) 14 (d) 18
Sol. (a) 7 +5 = 12
Case study based
Q.1 A state cricket authority has to choose a team of 11 members, to do it so the authority ask 2
coaches of a government academy to select the team
Members that have experience as well as best performer
in the last 15 matches. They can make up a team of 11
cricketers amongst 15 possible candidates in which
5 players can bowl.
(i) In how many ways can the final eleven be selected from 15 cricket players ?
(ii) In how many ways can the final eleven be selected if exactly 4 bowlers must be included.
(iii) In how many ways can the final eleven be selected if all bowlers must be included.
Sol. (i) 15C11= 1365
(ii) 4 bowlers can be select by = 5C4
Remaining 7 players can be select out of 10 (15-5) is = 10C7
Total no. of ways is = 5C4 .10C7 = 600
(iii) all 5 bowlers can be select by = 5C5
Remaining 6 players can be select out of 10 (15-5) is = 10C6
Total no. of ways is = 5C5 .10C6 = 210
Short answer type question
Q.1 How many number of rectangle forming by 5 different horizontal parallel lines and 7 other
different vertical parallel line.
Sol. A rectangle forming by two horizontal and two vertical parallel lines
No. of rectangle = 5C2 .7C2 = 10 . 21 = 210
Q.2 How many arrangement of the letters of the word APPLICATION.
Sol. Total letters 11, A = 2, P=2, I=2
11!
No. of arrangements = 2!. 2!. 2! = 4989600
Q.3There are 12 point on a circle. How many chord can be draw? Find number of intersection of
all chord in the circle?
Sol. No. of chord = 12C2
No. of intersection of chord = no. of quadrilateral can be made = 12C4
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MAIN CONCEPTS AND RESULTS
** Fundamental principle of counting, or( the multiplication principle): “If an event can occur in m
different ways, following which another event can occur in n different ways, then the total number of
occurrence of the events in the given order is m × n.”
**Factorial notation The notation n! represents the product of first n natural numbers, i.e., the product
1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this symbol as „n factorial‟.
Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !
n ! = n (n – 1) !
= n (n – 1) (n – 2) ! [provided (n ≥ 2)]
= n (n – 1) (n – 2) (n – 3) ! [provided (n ≥ 3)]
**Permutations A permutation is an arrangement in a definite order of a number of objects taken
some or all at a time.
** The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects
do not
repeat is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by
n!
P (n , r) OR n Pr ,0 r n
n r !
** n P0 1 n Pn
** The number of permutations of n different objects taken r at a time, where repetition is allowed, is
nr.
** The number of permutations of n objects, where p1 objects are of one kind, p2 are of second kind, ...,
n!
pk are of kth kind and the rest, if any, are of different kind is .
p1!p2!...pk !
** The number of permutations of an dissimilar things taken all at a time along a circle is ( n -1) !.
** The number of ways of arranging a distinct objects along a circle when clockwise and anticlockwise
1
arrangements are considered alike is (n -1) !.
2
** The number of ways in which (m + n) different things can be divided into two groups containing
( m n) !
m and n things is .
m ! n!
n!
Combination of n different objects taken r at a time, denoted by n C r .
r!n r !
** n Pr n Cr r! , 0 r n
* * n C0 1 n C n
n
C1 n n Cn 1
n n 1 n
n
C2 C2
2!
n n 1n 3 n
n
C3 Cn 3
3!
** n Cr n Cs r s or r s n
** n Cr n Cr 1 n 1Cr
44
, II. Some illustrations/Examples
MCQs
Q.1 How many three digit numbers are there with all distinct digits1,2,3,4,5,6,7,8,9,0 ?
(a) 458 (b) 568 (c) 648 (d) 748
Sol. (c)9.9.8= 648
9 9 8
Q.2The H.C.F. of 6!, 8!, 9!, 11!. is
(a) 6! (b) 8! (c) 9! (d) 11!
Sol. (a) 6!
Q.3 A polygon has 14 sides then number of its diagonals are:
(a) 91 (b) 77 (c) 28 (d) none of these
Sol. (b) 77
Let the no. of sides is n.
Then no. of its diagonals is = 14C2 – 14= 91 –14 = 77
Q.4 IfnC5 = nC7, find n.
(a) 12 (b) 15 (c) 14 (d) 18
Sol. (a) 7 +5 = 12
Case study based
Q.1 A state cricket authority has to choose a team of 11 members, to do it so the authority ask 2
coaches of a government academy to select the team
Members that have experience as well as best performer
in the last 15 matches. They can make up a team of 11
cricketers amongst 15 possible candidates in which
5 players can bowl.
(i) In how many ways can the final eleven be selected from 15 cricket players ?
(ii) In how many ways can the final eleven be selected if exactly 4 bowlers must be included.
(iii) In how many ways can the final eleven be selected if all bowlers must be included.
Sol. (i) 15C11= 1365
(ii) 4 bowlers can be select by = 5C4
Remaining 7 players can be select out of 10 (15-5) is = 10C7
Total no. of ways is = 5C4 .10C7 = 600
(iii) all 5 bowlers can be select by = 5C5
Remaining 6 players can be select out of 10 (15-5) is = 10C6
Total no. of ways is = 5C5 .10C6 = 210
Short answer type question
Q.1 How many number of rectangle forming by 5 different horizontal parallel lines and 7 other
different vertical parallel line.
Sol. A rectangle forming by two horizontal and two vertical parallel lines
No. of rectangle = 5C2 .7C2 = 10 . 21 = 210
Q.2 How many arrangement of the letters of the word APPLICATION.
Sol. Total letters 11, A = 2, P=2, I=2
11!
No. of arrangements = 2!. 2!. 2! = 4989600
Q.3There are 12 point on a circle. How many chord can be draw? Find number of intersection of
all chord in the circle?
Sol. No. of chord = 12C2
No. of intersection of chord = no. of quadrilateral can be made = 12C4
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