If $x =^{16} C_{5} +^{12} C_{4}, y = {\sum_{r = 1}^{3}}^{(20 - r)} C_{4}, {\sum_{k = 1}^{4}}^{(16 - k)} C_{3},$ then x + y + z =

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$19 \times 17 \times 45$

$19 \times 17 \times 15$

$19 \times 17 \times 16$

$19 \times 17 \times 48$

answer is D.

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Detailed Solution

$x = 16_{C_{5}} + 12_{C_{4}} y = \sum_{r = 1}^{3} {(20 - r)}_{C_{4}} z = \sum_{k = 1}^{4} {(10 - k)}_{C_{3}}$

$\begin{array}{rcl} x + y + z & = & 16_{c_{5}} + 12_{c_{4}} + 19_{C_{4}} + 18_{C_{4}} + 17_{C_{4}} + 15_{C_{3}} + 14_{C_{3}} + 13_{C_{3}} + 12_{C_{3}} \\ = & (12_{c_{3}} + 12_{C_{4}}) + 13_{C_{3}} + 14_{C_{3}} + 15_{C_{3}} + 16_{C_{5}} + 17_{C_{4}} + 18_{C_{4}} + 19_{C_{4}} \\ = & (13_{C_{4}} + 13_{C_{3}}) + 14_{C_{3}} + 15_{C_{3}} + 16_{C_{5}} + 17_{C_{4}} + 18_{C_{4}} + 19_{C_{4}} \\ = & 14_{C_{4}} + 14_{C_{3}} + 15_{C_{3}} + 16_{C_{5}} + 17_{C_{4}} + 18_{C_{4}} + 19_{C_{4}} \\ = & 15_{C_{4}} + 15_{C_{3}} + 16_{C_{5}} + 17_{C_{4}} + 18_{C_{4}} + 18_{C_{4}} + 19_{C_{4}} \\ = & 16_{C_{4}} + 16_{C_{5}} + 17_{C_{4}} + 18_{C_{4}} + 19_{C_{4}} \\ = & 17_{C_{5}} + 17_{C_{4}} + 18_{C_{4}} + 19_{C_{4}} \\ = & 18_{C_{5}} + 18_{C_{4}} + 19_{C_{4}} \\ = & 19_{C_{5}} + 19_{C_{4}} \\ = & 20_{C_{5}} = \frac{20 \times 19 \times + 18 \times 17 \times 16}{1 \times 2 \times 3 \times 4 \times 5} \\ = & 19 \times 17 \times 48 \end{array}$