# Permutation and Combination NEB CBSE[Class 12]

Questions and Solution of Combination

Q.1.Find the values of 14C5 , 10C8 and C(7,2).
Solution:
*14C5=
 14!
 5!(14−5)!

=
 14!
 5!9!

 14.13.12.11.10.9!
 5.4.3.2.1.9!

= 2002
*10C8=
 10!
 8!(10−8)!

=
 10!
 8!2!

 10.9.8!
 8!.2

= 45

* C(7,2)=
 7!
 2!(7−2)!

=
 7!
 5!2!

=
 7.6.5!
 5!.2

= 21

Q.2.Six friends want to play enough games of chess to be sure every one plays everyone else. How many games will they have to play?
Solution:
There are 6 players to be taken 2 at a time.
Using the formula:
C(6,2)=
 6!
 2!(6−2)!

=
 6×5×4!
 1×2×4!

=
 6×5
 2

=15
They will need to play 15 games.

Q.3.How many different committees of 4 students can be chosen from a group of 15?
Solution:
There are 15C4 possible combinations of 4 students from a set of 15.
15C4=
 15!
 4!(15−4)!

=
 15!
 4!×11!

=
 15×14×13×12×11!
 4!×11!

=
 15×14×13×12
 4×3×2×1

=1365

Q.4.In a group of 25 people, there are 15 Gents and 10 Ladies. Find the number of ways(combinations) of forming 15 people committee which should contain 9 Gents?
Solution:
Given: Total = 25 people, Gents = 12, Ladies = 10.
In 15 people committee, there are 9 Gents which means remaining 6 will be Ladies.
From 15 Gents, 9 Gents are selected. So the combination is C(15, 9).
From 10 Ladies, 6 Ladies are selected. So the combination is C(10, 6).
Therefore, Number of ways(combinations) can be calculated from the following formula:
= C(15, 9) * C(10, 6)
=
 15!
 9!(15−9)!
×
 10!
 6!(10−6)!

=
 15!
 9!×6!
×
 10!
 6!×4!

=5005×210
=10,51,050
Therefore, 10,51,050 combinations of Gents and Ladies are arranged

Q.5.If (1 + x)n = C0 + C1x + C2x2 + …… + Cnxⁿ,prove that C₀C₁+C₁Cₙ₋₂+...+CₙC₀=

.
Solution
(1 + x)n = C0 + C1x + C2x2 + …… + Cnxn …(i)
Also, (1 + x)n = Cnxn + Cn – 1.Xn – + …… + C1x + C0 ….(ii)
Multiplying (i) and (ii),
(1 + x)n.(1 + x)n = (C0 + C1x + C2x2 + …. + CnXn)(Cnxn + Cn – 1.xn – 1 + …. + C1x + C0)
(1 + x)2n. = C0Cnxn + C1.Cn – 1xn + …. + Cn.C0xn + ….
= (C0.Cn + C1.Cn – 1 + …… + Cn.C0)xn + …..  (iii).
Since (iii) is and identity, so the coefficient of xn of the L.H.S. should  be equal to the coefficient of xn of RHS.
Coeff. of xn In the expansion of (1 + x)2n = C(2n,n) …(iv)
Equating the coeff. of xn from (iii) and (iv),
C0.Cn + C1.Cn – 1  + …. + Cn.C0 = C(2n,n) =