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Combinations
Meaning of Combinations
 We know that the total number of permutations of
two persons out of four persons A, B, C, D is 4P2 =
12. They are as follows:
 AB AC AD
 BA BC BD
 CA CB CD
 DA DB DC
 Sometimes the persons are to be selected not
arranged. Two out of four persons A, B, C, D can be
selected in the following different ways:
 AB, AC, AD, BC, BD, CD
Meaning of Combinations
 Thus two persons out of four persons can be
selected in 6 different ways .
 These different ways of selection are known as
Combinations.
 AB and BA are two different permutations while in
combination the selection AB and BA is considered
only one.
 Notation: 4C2 = 6
Combination Formula
!
!( )!
n
r
n
C
r n r


( )
n n
r n rC C 
0 1n n
nC C 
1
1
n n n
r r rC C C
 
Ex.1 Find the values of the following:
1. 12C4
2. 25C23
3. 8C2
4. 11C7
Ex.2
 In how many ways a committee of 4 professors can
be formed out of 11 professors?
Solution:
 The total number of combinations of 4 professors out
of 11 professors
 = 11C4
 = 11! / ( 4! . 7!)
 =_____
 =330
Ex. 3
 In how many ways 4 Gujaratis, 2 Punjabis and 1
Madrasi can be selected out of 8 Gujaratis, 4
Punjabis and 3 Madarasis?
Solution:
 4 Gujaratis out of 8 can be selected in 8C4 ways.
 2 Punjabis out of 4 can be selected in 4C2 ways
 1 Madarasi out of 3 can be selected in 3C1 ways.
 Therefore total number of combinations
 = 8C4
x 4C2
x 3C1
 =______
 =1260
Restricted Combinations
 The number of combinations of n things taken r at a
time in which p particular things always occur is n-pCr-
p.
 p particular things never occur is n-pCr.
Ex. 4
 In how many ways a legal committee of 6 members
can be formed out of 11 ministers of a cabinet so
that the chief minister and the minister of law and
order are included?
 Moreover if two particular ministers are not to be
taken in the committee, in how many ways can it be
formed?
Solution:
 Total 6 ministers are to selected out of 11.
 If the chief minister and the minister of law and order
are to be included then remaining 4 ministers out of
9 can be selected in
 1C1 x 1C1 x 9C4=____
 =126
 If two particular ministers are not to be taken in
committee, then 1 C.M, 1 minister of law and order
and 4 out of 7 (9-2=7) ministers can be selected in
 1C1 x 1C1 x 7C4=_______
 =35
Ex.5
 A box contains 5 red, 3 green and 2 white balls. In
how many ways 3 balls can be drawn from it such
that
 (i) one ball of each colour is included?
 (ii) 2 balls of the same colour and 1 ball of different
colour is included?
 (iii) 3 balls of the same colour are included?
Solution:
 A box contains 5 red, 3 green and 2 white balls
 (i) the number of ways of drawing one ball of each of
the three colours
 = 5C1 x 3C1 x 2C1 = _______
 =30
 (ii) 2 balls of the same colour and 1 ball of different
colour can be draws in the following ways:
 2 red balls and 1 from the remaining 5 balls OR
 2 green balls and 1 from the remaining 7 balls OR
 2 white balls and 1 from the remaining 8 balls
 Therefore, the total number of combinations +
 = 5C2 x 5C1 + 3C2 x 7C1 + 2C2 x 8C1
 =________
 =79
 (iii) 3 balls of the same colour can be all the three
red balls or all the three green balls
 Therefore the total number of ways
 = 5C3 + 3C3
 =___
 =11
Ex.6
 A cricket team consists of 15 players including 4
bowlers and 2 wicket keepers. In how many ways 11
players can be selected of them so that 3 bowlers
and 1 wicket-keeper are included in the team?
Solution:
 Total 11 players are to be selected from 15.
 3 bowlers out of 4, 1 wicket-keeper out of 2 and
other 7 players from the remaining 9 can be selected
in
 = 4C3 x 2C1 x 9C7 = _______
 =288

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Combinations

  • 2. Meaning of Combinations  We know that the total number of permutations of two persons out of four persons A, B, C, D is 4P2 = 12. They are as follows:  AB AC AD  BA BC BD  CA CB CD  DA DB DC  Sometimes the persons are to be selected not arranged. Two out of four persons A, B, C, D can be selected in the following different ways:  AB, AC, AD, BC, BD, CD
  • 3. Meaning of Combinations  Thus two persons out of four persons can be selected in 6 different ways .  These different ways of selection are known as Combinations.  AB and BA are two different permutations while in combination the selection AB and BA is considered only one.  Notation: 4C2 = 6
  • 4. Combination Formula ! !( )! n r n C r n r   ( ) n n r n rC C  0 1n n nC C  1 1 n n n r r rC C C  
  • 5. Ex.1 Find the values of the following: 1. 12C4 2. 25C23 3. 8C2 4. 11C7
  • 6. Ex.2  In how many ways a committee of 4 professors can be formed out of 11 professors?
  • 7. Solution:  The total number of combinations of 4 professors out of 11 professors  = 11C4  = 11! / ( 4! . 7!)  =_____  =330
  • 8. Ex. 3  In how many ways 4 Gujaratis, 2 Punjabis and 1 Madrasi can be selected out of 8 Gujaratis, 4 Punjabis and 3 Madarasis?
  • 9. Solution:  4 Gujaratis out of 8 can be selected in 8C4 ways.  2 Punjabis out of 4 can be selected in 4C2 ways  1 Madarasi out of 3 can be selected in 3C1 ways.  Therefore total number of combinations  = 8C4 x 4C2 x 3C1  =______  =1260
  • 10. Restricted Combinations  The number of combinations of n things taken r at a time in which p particular things always occur is n-pCr- p.  p particular things never occur is n-pCr.
  • 11. Ex. 4  In how many ways a legal committee of 6 members can be formed out of 11 ministers of a cabinet so that the chief minister and the minister of law and order are included?  Moreover if two particular ministers are not to be taken in the committee, in how many ways can it be formed?
  • 12. Solution:  Total 6 ministers are to selected out of 11.  If the chief minister and the minister of law and order are to be included then remaining 4 ministers out of 9 can be selected in  1C1 x 1C1 x 9C4=____  =126  If two particular ministers are not to be taken in committee, then 1 C.M, 1 minister of law and order and 4 out of 7 (9-2=7) ministers can be selected in  1C1 x 1C1 x 7C4=_______  =35
  • 13. Ex.5  A box contains 5 red, 3 green and 2 white balls. In how many ways 3 balls can be drawn from it such that  (i) one ball of each colour is included?  (ii) 2 balls of the same colour and 1 ball of different colour is included?  (iii) 3 balls of the same colour are included?
  • 14. Solution:  A box contains 5 red, 3 green and 2 white balls  (i) the number of ways of drawing one ball of each of the three colours  = 5C1 x 3C1 x 2C1 = _______  =30  (ii) 2 balls of the same colour and 1 ball of different colour can be draws in the following ways:  2 red balls and 1 from the remaining 5 balls OR  2 green balls and 1 from the remaining 7 balls OR  2 white balls and 1 from the remaining 8 balls
  • 15.  Therefore, the total number of combinations +  = 5C2 x 5C1 + 3C2 x 7C1 + 2C2 x 8C1  =________  =79  (iii) 3 balls of the same colour can be all the three red balls or all the three green balls  Therefore the total number of ways  = 5C3 + 3C3  =___  =11
  • 16. Ex.6  A cricket team consists of 15 players including 4 bowlers and 2 wicket keepers. In how many ways 11 players can be selected of them so that 3 bowlers and 1 wicket-keeper are included in the team?
  • 17. Solution:  Total 11 players are to be selected from 15.  3 bowlers out of 4, 1 wicket-keeper out of 2 and other 7 players from the remaining 9 can be selected in  = 4C3 x 2C1 x 9C7 = _______  =288