Question

# In a shop, there are five types of ice-creams available. A child buys six ice-creams.Statement-l: 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.A. Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1B. Statement-1 is true, Statement-2 is false.C. Statement-1 is false, Statement-2 is true.D. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hint: The number of different ways the child can buy 6 ice-creams from five types of ice-creams can be calculated by the concept of non-integral solutions. The number of non-negative integral solutions of equation ${x_1} + {x_2} + {x_3} + ... + {x_r} = n$ is ${}^{(n + r - 1)}{C_{(r - 1)}}$ . If out of n things, p are exactly of one kind and q exactly alike, then the number of ways of arranging n things taken all at a time $= \dfrac{{n!}}{{p!q!}}$

The number of different ways the child can buy 6 ice-creams from five types of ice-creams
= Number of non- negative integral solutions of the equation ${x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 6$
Since, the number of positive integral solutions of equation ${x_1} + {x_2} + {x_3} + ... + {x_r} = n$ is ${}^{(n + r - 1)}{C_{(r - 1)}}$ . Here, n=6 and r=5
Therefore, the number of different ways the child can buy 6 ice-creams will be,
$\Rightarrow {}^{6 + 5 - 1}{C_{5 - 1}} = {}^{10}{C_4}$
Now, as we know out of n things, p are exactly of one kind and q exactly alike, then the number of ways of arranging n things have taken all at a time $= \dfrac{{n!}}{{p!q!}}$.
Here, n is the sum of 6 A’s and 4 B’s. Therefore, n is 10, p is 6 and q is 4.
$\Rightarrow \dfrac{{10!}}{{6!4!}} = {}^{10}{C_4}$
Hence, Statement-2 is True.

So, the correct answer is “Option C”.

Note: Additional information, the Number of Non-negative integral solutions. Consider the equation, ${x_1} + {x_2} + {x_3} + ... + {x_r} = n$. In this case, $\left( {{x_1},{x_2},...,{x_n}} \right) \geqslant 0$. Therefore, $\left( {{x_1} + 1,{x_2} + 1,...,{x_n} + 1} \right) \geqslant 1$. Substitute ${x_1} + 1 = {y_1},{x_2} + 1 = {y_2}$ and similarly to ${y_n}$ in the given equation such that equation becomes ${y_1} + {y_2} + {y_3} + ... + {y_r} = n + r$
Therefore, Number of non-negative integral solutions of equation ${x_1} + {x_2} + {x_3} + ... + {x_r} = n$ is ${}^{(n + r - 1)}{C_{(r - 1)}}$