Note : n items can be distribute among p persons are $${}^{n + p - 1}{C_{p - 1}}$$ ways.

Here n = 6 ice-cream

p = 5 types of ice-cream

Each ice-cream belongs to one of the 5 ice-cream type. So chosen 6 ice-crean can be divide into 5 types of ice-cream.

$$ \therefore $$ The number of different ways the child can buy the six ice-cream is = $${}^{6 + 5 - 1}{C_{5 - 1}}$$ = $${}^{10}{C_4}$$

$$ \therefore $$ Statement - 1 is false.

Number of different ways of arranging 6 A and 4 B's in a row

= $${{10!} \over {6!4!}} = {}^{10}{C_4}$$

$$ \therefore $$ Statement - 2 is true.

The total number of ways is

$${}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}$$

A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.

Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$

Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$

Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$

Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$
So, total no of ways he can give votes
= $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$
= 385

Note : Here we use addition rule as he can vote any one of those four rules. Whenever there is "or" choices, we use addition rule.