If x=16C5+12C4,y=∑r=1320−rC4 and z=∑k=1416−kC3 then x+y+z =

x+y+z=(16C5+12C4)+(19C4+18C4+17C4) +(15C3+14C3+13C3+12C3) =(12C4+16C5)+(17C4+18C4+19C4) +(12C3+13C3+14C3+15C3) =13C4+16C5+17C4+18C4+19C4+13C3+14C3 +15C3 =14C4+16C5+17C4+18C4+19C4+14C3+15C3 =15C4+16C5+17C4+18C4+19C4 =16C4+16C5+17C4+18C4+19C4 =17C5+17C4+18C4+19C4 =18C5+18C4+19C4 =19C5+19C4 =20C5 =20!5!15! =19×48×17

If x=16C5+12C4,y=∑r=1320−rC4 and z=∑k=1416−kC3 then x+y+z =

x+y+z=(16C5+12C4)+(19C4+18C4+17C4) +(15C3+14C3+13C3+12C3) =(12C4+16C5)+(17C4+18C4+19C4) +(12C3+13C3+14C3+15C3) =13C4+16C5+17C4+18C4+19C4+13C3+14C3 +15C3 =14C4+16C5+17C4+18C4+19C4+14C3+15C3 =15C4+16C5+17C4+18C4+19C4 =16C4+16C5+17C4+18C4+19C4 =17C5+17C4+18C4+19C4 =18C5+18C4+19C4 =19C5+19C4 =20C5 =20!5!15! =19×48×17

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

$\begin{aligned} x + y + z = (^{16} C_{5} +^{12} C_{4}) + (^{19} C_{4} +^{18} C_{4} +^{17} C_{4}) \\ + (^{15} C_{3} +^{14} C_{3} +^{13} C_{3} +^{12} C_{3}) \\ = (^{12} C_{4} +^{16} C_{5}) + (^{17} C_{4} +^{18} C_{4} +^{19} C_{4}) \\ + (^{12} C_{3} +^{13} C_{3} +^{14} C_{3} +^{15} C_{3}) \\ =^{13} C_{4} +^{16} C_{5} +^{17} C_{4} +^{18} C_{4} +^{19} C_{4} +^{13} C_{3} +^{14} C_{3} \end{aligned}$
$\begin{aligned} +^{15} C_{3} \\ =^{14} C_{4} +^{16} C_{5} +^{17} C_{4} +^{18} C_{4} +^{19} C_{4} +^{14} C_{3} +^{15} C_{3} \\ =^{15} C_{4} +^{16} C_{5} +^{17} C_{4} +^{18} C_{4} +^{19} C_{4} \\ =^{16} C_{4} +^{16} C_{5} +^{17} C_{4} +^{18} C_{4} +^{19} C_{4} \\ =^{17} C_{5} +^{17} C_{4} +^{18} C_{4} +^{19} C_{4} \\ =^{18} C_{5} +^{18} C_{4} +^{19} C_{4} \\ =^{19} C_{5} +^{19} C_{4} \\ =^{20} C_{5} \\ = \frac{20!}{5! 15!} \\ = 19 \times 48 \times 17 \end{aligned}$