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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-1

B. 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.

Answer
Verified

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.

Therefore, Number of non-negative integral solutions of equation ${x_1} + {x_2} + {x_3} + ... + {x_r} = n$ is ${}^{(n + r - 1)}{C_{(r - 1)}}$

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