Beggar’s Method

The principle of multiplication states that “If we can perform a particular operation in ‘n’ ways and the second operation in ‘m’ ways, the two operations can be performed in m x n ways in succession”. This is applicable to a finite number of operations. A factorial can be defined as a function that multiplies a single number with each and every number preceding it. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120. The concept of permutations is the arrangement of objects in a particular order. The formula for permutation is given by:

ⁿP_r = n! / [n – r]! (without repetition)

ⁿP_r = n! / [p! q! r!]

Example: Find the number of ways in which 5 prizes can be distributed among 4 boys where every boy can take one or more prizes.

Solution:

The 1^st prize can be distributed to any of the 4 boys, hence it is done in 4 ways. In the same way, the second, third, fourth and fifth prizes can be given in 4 ways. Total number of ways = 4 * 4 * 4 * 4 * 4 = 4⁵ = 1024 ways.

Combination refers to the selection of objects without repetition where the order doesn’t matter. The formula for the combination of n things being chosen out of r is given by:

ⁿC_r = [n!] / [n – r]! r!

Beggar’s Method

It is based on the distribution of like objects. It states that “The number of ways of distributing ‘n’ identical things among ‘p’ persons without any restriction (none, 1, 2 or all or any of the number of things can be given to one person)” = ^n+p-1C_p-1.

r₁ + r₂ + …… + r_p = n

The coefficient of xⁿ in (1 + x + …… + xⁿ)^p

= [(1 – xⁿ⁺¹) / (1 – x)]^p

= (1 – xⁿ⁺¹)^p (1 – x)^p

The coefficient of xⁿ in (1 – x)^-p

= ^p+n-1C_n

= ^n+p-1C_p-1

Illustration 1: In how many ways can 3 rings be worn on 4 fingers if any number of rings can be worn on any finger?

(i) Rings are distinct

(ii) Rings are identical

Solution:

(i) Rings are distinct

Let R₁, R₂ and R₃ be the rings.

Number of ways = 4³

= 64

(ii) Rings are identical

Here, n = 3, p = 4

Using the formula from the Beggar’s method, ^n+p-1C_p-1 = ^3+4-1C_4-1

= ⁶C₃

= 20

Illustration 2: Find the number of ways of distributing 10 apples, 5 mangoes, and 4 oranges among 4 persons if each can receive any number of fruits and the same type of fruits is identical.

Solution:

Here, p = 4

Using the formula from the Beggar’s method, ^n+p-1C_p-1

= [^10+4-1C_4-1] (apples) [^5+4-1C_4-1] (mangoes) [^4+4-1C_4-1] (oranges)

= ¹³C₃ ⁸C₃ ⁷C₃

Illustration 3: Find the number of ways in which 16 identical toys are distributed among 3 students such that each receives not less than 3 toys.

Solution:

Let the students be S₁, S₂, and S₃ such that each receives not less than 3 toys.

S₁ + S₂ + S₃ = 16 —- (1)

Distribute 3 toys to each of the students in the beginning.

So, equation (1) now becomes S₁’ + S₂’ + S₃’ = 16 – 9 = 7

Using the formula from the Beggar’s method, ^n+p-1C_p-1

= ^7+3-1C_3-1

= ⁹C₂

Beggar’s Method – Video Lesson