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(A) The number of ways to buy $6$ ice creams from $5$ types (with repetition allowed) is given by the formula $^{n+r-1}C_{r}$,where $n=5$ and $r=6$.Th

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Statement-$1$ is false,Statement-$2$ is true.

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(A) The number of ways to buy $6$ ice creams from $5$ types (with repetition allowed) is given by the formula $^{n+r-1}C_{r}$,where $n=5$ and $r=6$.This equals $^{5+6-1}C_{6} = ^{10}C_6 = ^{10}C_4 = 210$.Statement-$1$ says $^{10}C_5$,which is $252$. Thus,Statement-$1$ is false.The number of ways to arrange $6$ '$A$'s and $4$ '$B$'s in a row is given by the permutation of multisets formula: $\frac{(6+4)!}{6! 	imes 4!} = \frac{10!}{6! 	imes 4!} = ^{10}C_4 = 210$.This matches the calculation for the number of ways to buy the ice creams.Therefore,Statement-$2$ is true.

There are $5$ types of ice creams available in a shop. $A$ child buys $6$ ice creams.
Statement-$1$: The child can buy $6$ ice creams in $^{10}C_5$ ways.
Statement-$2$: The number of ways the child can buy $6$ ice creams is equal to the number of ways to arrange $6$ '$A$'s and $4$ '$B$'s in a row.

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There are $5$ types of ice creams available in a shop. $A$ child buys $6$ ice creams.Statement-$1$: The child can buy $6$ ice creams in $^{10}C_5$ ways.Statement-$2$: The number of ways the child can buy $6$ ice creams is equal to the number of ways to arrange $6$ '$A$'s and $4$ '$B$'s in a row.