The formal expression of binomial theorem is as follows:
⦿ (1 + x)ⁿ = 1 + ⁿC₁ x +ⁿC₂ x² + ⁿC₃ x³ + ⁿC₄ x⁴ + ……….
now understand multinomial theorem:-
∴ Coefficient of xʳ in (1-x)⁻ⁿ
= ⁿ ⁺ ʳ ⁻¹Cᵣ
& Since we know that ⁿCᵣ = ⁿCₙ₋ᵣ
∴ ⁿ ⁺ ʳ ⁻¹Cᵣ = ⁿ ⁺ ʳ ⁻¹C₍ₙ ₊ ᵣ ₋ ₁₎ – ᵣ
= ⁿ ⁺ ʳ ⁻¹Cₙ₋₁
Example:-
⦿ Coefficient of x¹⁰ in (1-x)⁻ⁿ = ¹⁰⁺⁽ⁿ⁻¹⁾Cₙ₋₁
⦿ Coefficient of x¹⁵ (1-x)⁻ⁿ = ¹⁵⁺ⁿ⁻¹Cₙ₋₁
⦿ Coefficient of x⁸ in (1-x)⁻ⁿ = ⁸⁺⁽⁵⁻¹⁾C₅₋₁ = ¹²C₄
(Q). Find the coefficient of x⁶ in ((1+x)¹⁰ + 10x³ + 6x⁶)
⦿ n identical objects can be distributed among r person in
So distribution of n identical object among r person & number of integral solution of equation x₁ + x₂ + x₃ + x₄ + x₅ + ……….. + xᵣ = n
where x₁, x₂, x₃, x₄, x₅,……….xᵣ ≥ 0
have the same solution which is
Using Multinomial theorem to find the number of non-negative integral solution of equation
x₁ + x₂ + x₃ + x₄ + x₅ + ……….. + xᵣ = n
(Q). Find the number of non-negative integral solution of x + y + z = 20
Solution-
x + y + z = 20 x≥0
y≥0
z≥0
= ²⁰⁺²C₂
= ²²C₂ Answer
method(2):S= coefficient of p²⁰ in (p⁰ + p¹ + p² + …….. p²⁰)³
= coefficient of p²⁰ in (1 + p¹ + p² + …….. p²⁰ + p²¹ + p²² + …….. ∞)³
= coefficient of p²⁰ in
= coefficient of p²⁰ in (1-p)⁻³
= ²⁰ ⁺ ⁽³⁻¹⁾C₃₋₁
=²²C₂ Answer
(Q). Find number of positive integral solution of x + y + z = 20
Solution:-
x + y + z = 20 x≥ 1
y≥ 1
z≥1
Method (2):-
Coefficient of p²⁰ in (p¹ + p² + p³ …….. p²⁰)³
= Coefficient of p²⁰ in (p + p² + p³ + …….. p²⁰+ ……..∞)³
= Coefficient of p²⁰ in
= Coefficient of p²⁰ in p³(1-p)⁻³
= Coefficient of p¹⁷ in (1-p)⁻³
= ¹⁷⁺⁽³⁻¹⁾C₍₃₋₁₎
= ¹⁹C₂ Answer
(Q). Find the number of non-negative integral solution of a + b + c + 4d = 20
Solution:-
a + b + c + 4d = 20 20≥a≥0
20≥b≥0
20≥b≥0
5≥b≥0
Method(1):
a + b + c + 4d = 20
⦿ d = 0 ⟹ a + b + c = 20 ⟹ ²²C₂
⦿ d = 1 ⟹ a + b + c = 16 ⟹ ¹⁸C₂
⦿ d = 2 ⟹ a + b + c = 12 ⟹ ¹⁴C₂
⦿ d = 3 ⟹ a + b + c = 8 ⟹ ¹⁰C₂
⦿ d = 4 ⟹ a + b + c = 4 ⟹ ⁶C₂
⦿ d = 5 ⟹ a + b + c = 0 ⟹ ²C₂
Hence Answer = ²²C₂ + ¹⁸C₂ + ¹⁴C₂
+ ¹⁰C₂ + ⁶C₂ + ²C₂
= 231 + 153 + 91 + 45 + 15 + 1
=536 Answer
Method(2):
a + b + c + 4d = 20
⟹ a + b + c = 20 – 4d
Let d be a contant k where 5≥k≥0
∴ a + b + c = 20 – 4k 5≥k≥0
= ²²C₂ + ¹⁸C₂ +¹⁴C₂ + ¹⁰C₂ + ⁶C₂ + ²C₂
= 536 Answer
Methode(3): Using multinomial theorem:-
Coefficient of p²⁰ in (p⁰ + p¹ + p² + p³ …….. + p²⁰)³(p⁰ + p⁴ + p⁸ + p¹² + p¹⁶ + p²⁰)
= Coefficient of p²⁰ in (1 + p + p² + p³ …….. + p²⁰ + ……….∞)³ (p⁰ + p⁴ + p⁸ + p¹² + p¹⁶ + p²⁰)
= Coefficient of p²⁰ in (p⁰ + p⁴ + p⁸ + p¹² + p¹⁶ + p²⁰)
= Coefficient of p²⁰ in (1-p)⁻³ (p⁰ + p⁴ + p⁸ + p¹² + p¹⁶ + p²⁰)
= ²²C₂ + ¹⁸C₂ + ¹⁴C₂ + ¹⁰C₂ + ⁶C₂ + 1
= 536 Answer
(Q). Find number of non-negative integral solution of inequality
x₁ + x₂ + x₃ + x₄ + x₅ ≤ 20
Solution:-
x₁ + x₂ + x₃ + x₄ + x₅ + k = 20 where 20≥k≥0
when k=0 then x₁ + x₂ + x₃ + x₄ + x₅ = 20
when k=1 then x₁ + x₂ + x₃ + x₄ + x₅ = 19
when k=2 then x₁ + x₂ + x₃ + x₄ + x₅ =18
⋮ ⋮
⋮ ⋮
⋮ ⋮
⋮ ⋮
when k=20 then x₁ + x₂ + x₃ + x₄ + x₅ = 0
So on depending value of k (20≥k≥0)
value of expression automatically becomes
x₁ + x₂ + x₃ + x₄ + x₅ ≤ 20
⇓
Hence number of intergral solution = ⁵⁺²⁰C₅ = ²⁵C₅ Answer
(Q). Find Number of integral solution off equation
x₁ + x₂ + x₃ + x₄ = 20
;where x₁≥2
x₂≥4
x₃≥0
x₄≥6
Solution:-
Method(1):
Method(2): Using Multinomial theorem:-
Coefficient of p²⁰ in (p² + p³ + p⁴ + ……… + p²⁰)(p⁴ + p⁵ + p⁶ + ……… + p²⁰)(p⁰ + p¹ + p² + p³ + …….. + p²⁰)(p⁶ + p⁷ + p⁸ + ….. + p²⁰)
= Coefficient of p²⁰ in (p² + p³ + p⁴ + …….. + ∞)(p⁴ + p⁵ + p⁶ + ……… ∞)(p⁰ + p¹ + p² + p³ + …….. ∞)(p⁶ + p⁷ + p⁸ + ….. ∞)
= Coefficient of p²⁰ in
=Coefficient of p²⁰ in p¹²(1-p)⁻⁴
= Coefficient of p⁸ in (1-p)⁻⁴
=⁸⁺³C₃
= ¹¹C₃ Answer
(Q). Find number of non-negative integral solution of
a + b + c + d + e + f = 30
a + b + c = 17
Solution:-
(Q). There are two sections A & B and there are 16 intermediate stations between A & B. A train is going from station A toward station B. In how many ways train can stop at 4 stations if no two stations should be consecutive. Train can take stoppage at any of these A, B & 16 intermediate stations.
Solution:-
(Q). There are 18 chairs. These chairs are to be occupied by 4 students. Find the number of possible arrangement if:-
(i) No two students sit side-by-side.
(ii) There should be atleast 3 empty chairs between any two students.
Solution:-
(i)
(Q) There are unlimited number of balls of colours Red, Green, Blue & White. They are all alike except for colour. In how many ways 20 balls can be selected?
Solution:-
(Q). Let a student can score maximum 100 marks in physics, chemistry & math each. Then find the number of ways in which he can score a total of 170 marks while getting atleast 50 marks in each subjects. Only integral marks are allowed.
Solution:-
P + C + M = 170
100≥P≥50
100≥C≥50
100≥M≥50
Using Multinomial theorem
(Q). In the above question if the total numbers (marks) required are 240.
Solution:-
Coefficient of x²⁴⁰ in
= Coefficient of x²⁴⁰ in x¹⁵⁰(1-x⁵¹)³(1-x)⁻³
= Coefficient of x⁹⁰ in (1-x⁵¹)³(1-x)⁻³
= Coefficient of x⁹⁰ in (1 – 3x⁵¹ + 3x¹⁰² – x¹⁵³)(1-x)⁻³
✅ ✅
= ⁹⁹C₂ – 3.⁴¹C₂ Answer
(Q). In how many ways a batman can score 20 runs out of six balls if he can score 0, 1, 2, 3, 4, 5 or 6 runs at a particular ball?
Solution:-
x₁ + x₂ + x₃ + x₄ + x₅ + x₆ = 20
6≥x₁, x₂, x₃, x₄, x₅, x₆≥0
∴ Coefficient of p²⁰ in (p⁰ + p¹ + p² + p³ + p⁴ + p⁵ + p⁶)⁶
= Coefficient of p²⁰ in
= Coefficient of p²⁰ in (1 – p⁷)⁶(1 – p)⁻⁶
= ²⁵C₅ – ⁶C₁×¹⁸C₅ + ⁶C₂×¹¹C₅ Answer
(Q). a+b+c+d+e+f = 20; find the number of integral solution of this equation
If 10≥a≥5
b, c, d, e≥0
Solution:-
Method (2):
Coefficient of p²⁰ in (p⁵ + p⁶ + p⁷ + p⁸)( p⁷ + p⁸ + p⁹ + p10)(p⁰ + p¹ + p² + p³ + ……. + p²⁰)⁴
= Coefficient of p²⁰ in p⁵(1 + p + p² + p³).p⁷.(1 + p + p² + p³)(1 + p + p² + p³+ …….. p²⁰)⁴
= Coefficient of p²⁰ in p¹²(1 + p + p² + p³)²((1 + p + p² + p³ + ……… + p²⁰)⁴
= Coefficient of p⁸ in
= Coefficient of p⁸ in (1-p⁴)²(1-p)⁻⁶
= Coefficient of p⁸ in (1-2p⁴+p⁸)(1-p)⁻⁶
= ¹³C₅ – 2×⁹C₅ + 1
= 1036 Answer
(Q). Find number of integral solution of equation
x₁ + x₂ + x₃ + x₄ + x₅ = 48
if x₁, x₂, x₃, x₄, x₅ ← evennumbers; ≥0
Solution:-
2n₁ + 2n₂ + 2n₃ + 2n₄ + 2n₅ = 48
⟹ n₁ + n₂ + n₃ + n₄ + n₅ = 24
≥0 ≥0 ≥0 ≥0 ≥0
= ²⁸C₄ Answer