Let binom{n}{k} denote {}^{n}C_{k} and left[b | vedclass

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(D) Using the identity $\sum_{i=0}^{r} \binom{n}{i} \binom{m}{k-i} = \binom{n+m}{k}$,we simplify $A_{k}$.$A_{k} = \sum_{i=0}^{9} \binom{9}{i} \binom{1

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$49$

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(D) Using the identity $\sum_{i=0}^{r} \binom{n}{i} \binom{m}{k-i} = \binom{n+m}{k}$,we simplify $A_{k}$.$A_{k} = \sum_{i=0}^{9} \binom{9}{i} \binom{12}{k-i} + \sum_{i=0}^{8} \binom{8}{i} \binom{13}{k-i}$.Applying Vandermonde's Identity:$A_{k} = \binom{9+12}{k} + \binom{8+13}{k} = \binom{21}{k} + \binom{21}{k} = 2 \binom{21}{k}$.Now,calculate $A_{4} - A_{3} = 2 \left( \binom{21}{4} - \binom{21}{3} ight)$.$\binom{21}{4} = \frac{21 	imes 20 	imes 19 	imes 18}{4 	imes 3 	imes 2 	imes 1} = 5985$.$\binom{21}{3} = \frac{21 	imes 20 	imes 19}{3 	imes 2 	imes 1} = 1330$.$A_{4} - A_{3} = 2(5985 - 1330) = 2(4655) = 9310$.Given $190p = 9310$,we find $p = \frac{9310}{190} = 49$.

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