Concept:

Suppose a set of n objects has n1 of one kind of object, n2 of a second kind, n3 of a third kind, and so on with n = n1 + n2 + n3  +...+ nk then the number of distinguishable permutations of the n objects is  = \(\frac{{{\rm{n}}!}}{{{{\rm{n}}_1}! \times {{\rm{n}}_2}! \times {{\rm{n}}_3}! \ldots \ldots \ldots {{\rm{n}}_{\rm{k}}}!}}\)

Addition and Multiplication principal:

• If there are m ways to choose an object one and n ways to choose an object two then the number of ways of selecting objects one and two are given by m × n.
• If there are m ways to choose an object one and n ways to choose an object two then the number of ways of selecting objects one or two is given by m + n.

Calculation:

Total no. of toys = 5
Total no. of children = 3
Each should get one toy
Selection can be done as follows: (2, 2, 1) or (1, 1, 3)

= ⁵C₂ × ³C₂ × ¹C₁ × \(\frac {3!}{2!}\) + ⁵C₁ × ⁴C₁ × ³C₃ × \(\frac {3!}{2!}\) 

= (10 × 3 × 1  × 3) +  (5 × 4 × 1 × 3)

= 90 + 60

= 150