In a shop there are five types of ice-creams available. A child buys six ice-creams. Statement-1: The number of different ways the child can buy the six ice-creams is $${}^{10}C_5$$. Statement-2: The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging $$6\ A's$$ and $$4\ B's$$ in a row.

Let the five types of ice-creams be denoted by the variables $$x_1,x_2,x_3,x_4,x_5$$, where $$x_i$$ counts how many ice-creams of the $$i^{\text{th}}$$ type are bought.

The child buys six ice-creams in all, so we need the number of non-negative integer solutions of
$$x_1+x_2+x_3+x_4+x_5=6.$$

The standard “stars and bars’’ formula for the number of solutions of
$$x_1+x_2+\dots+x_k=n$$ with each $$x_i\ge 0$$ is $$\binom{n+k-1}{k-1}.$$
Here $$n=6$$ and $$k=5$$, therefore the required count is
$$\binom{6+5-1}{5-1}= \binom{10}{4}.$$

$$\binom{10}{4}=\binom{10}{6}\neq\binom{10}{5},$$ so Statement-1, which claims $$\binom{10}{5},$$ is false.

Now consider Statement-2. Arranging $$6$$ identical A’s and $$4$$ identical B’s in a row produces a string of length $$10$$ containing exactly $$6$$ A’s. To form such a string we only have to choose the positions of the $$4$$ B’s (or, equivalently, the $$6$$ A’s). Hence the number of distinct arrangements is
$$\binom{10}{4}=\binom{10}{6}.$$

This matches the correct count for buying six ice-creams, so Statement-2 is true.

Therefore: Option A which is: Statement-1 is false, Statement-2 is true.