**Multinomial Theorem**

Consider the equation x₁ + x₂ + … + xᵣ = n, where aᵢ ≤ xᵢ ≤ bᵢ; xᵢ ϵ 1; I = 1, 2, …, r.

In order to find the number of solutions of the given equation satisfying the given condition. We observe that the number of solutions is the same as the coefficient of xⁿ in the product.

(x^{a₁}
+ x^{a₁}⁺¹ + x^{a₁}⁺² + … + x^{b₁}) x (x^{a₂}
+ x^{a₂}⁺¹ + x^{a₂}⁺² + … + x^{b₂}) x (x^{a₃}
+ x^{a₃}⁺¹ + x^{a₃}⁺² + … + x^{b₃}) x … x (x^{aᵣ}
+ x^{aᵣ}⁺¹ + x^{aᵣ}⁺² + … + x^{bᵣ})

For example, if we have to find the number of non-negative integral solution of x₁ + x₂ + … + xᵣ = n, then as above, the required number is the coefficient of xⁿ in (xᴼ + x¹ + … + xⁿ) (xᴼ + x¹ + … + xⁿ) (xᴼ + x¹ + … + xⁿ) … (xᴼ + x¹ + … + xⁿ) (r – brackets)

= coefficient of xⁿ in (1 + x + x² + …+ xⁿ)ʳ

= coefficient of xⁿ in (1 + x + …)ʳ

= coefficient of xⁿ in (1 – x)ʳ

= coefficient of xⁿ in \(\left( 1+(-r)(-x)+\frac{(-r)(-r-1)}{2!}{{(-x)}^{2}}+\frac{(-r)(r-1)(-r-2)}{3!}{{(-x)}^{3}} \right)\).

= coefficient of xⁿ in [1 + ʳC₁x + ʳ⁺¹C₂x² + …]

= ⁿ⁺ʳ⁻¹C_{n}

= ⁿ⁺ʳ⁻¹Cᵣ₋₁

**Example:** In how many ways the sum of upper
faces of four distinct dices can be six

**Solution**: Here, the number of required ways
will be equal to the number of solutions of x₁ + x₂ + x₃ + x₄ = 6, i.e., 1 ≤
x₁, x₂, x₃, x₄ ≤ 6.

∵ The upper limit is six, which is equal to the sum required.

So, the upper limit taken as infinite. So, number of solution is equal to the coefficient of t⁵ (1+ t + t² + …)⁴

= coefficient of t⁵ in (1 – t)⁻⁴

= ^{(}⁶⁺⁴⁻¹^{)}C_{(}₄₋₁_{)}

= ⁹C₃ = 84.