Mr. & Mrs Paul have 'a' number of children. The children wanted to gift something to their parents on their marriage anniversary. They shortlisted 'b' types of gifts and each child bought one gift for the parents. If each child thought of buying a different gift, what is the probability that, even after trying their best to buy different gifts, all of them bought the same gift, given b > a?

A. \(\frac{1}{b}^{a}\)

B. \(\frac{1}{b}^{(a-1)}\)

C. \(\frac{a}{b^{a}}\)

D. 1 - \(\frac{1}{b}^{(a-1)}\)

E. 1 - \(\frac{1}{b}^{a}\)

Why is it equal to b^a and not simply b*a?

If i have 3 people and 4 catagories, my total is 3*4 combinations.
a1b1
a1b2
a1b3
a1b4
...
a3b3
a3b4
totaling to 12.

4^3 doesn't equal 12.

It looks like you're treating this as one co-dependent action not separate independent actions, and i'm not sure why.
The actions of one person picking a gift type does not influence the action of the next person picking a gift type, so the combinations should be independent and should be added, not multiplied.

Please explain why this situation is 4C1*4C1*4C1 and not 4C1+4C1+4C1.

Edit:
Nevermind - i understand my thought process mistake.
Adding would give you the selection of 1 kid OR another kid selecting a gift, not 1 kid AND another kid selecting a gift
a1b1 a2b1
a1b1 a2b2

they are actually codependent.

Say a = 3 and b = 10.
The first kid can choose from 10 gifts, the second can choose from 10, the third can choose from 10.
There are 10 choices being made thrice; 10*10*10 = 10^(3) -> b^(a).