If x={ }^{16} C_5+{ }^{12} C_4, y=sum_{r=1}^3 | vedclass

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(D) Given $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$. Then $x+y+z={ }^{16} C_5+{ }^{12} C_4+\left({

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$19	imes 17	imes 48$

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(D) Given $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$. Then $x+y+z={ }^{16} C_5+{ }^{12} C_4+\left({ }^{19} C_4+{ }^{18} C_4+{ }^{17} C_4ight)+\left({ }^{15} C_3+{ }^{14} C_3+{ }^{13} C_3+{ }^{12} C_3ight)$. Rearranging the terms: $x+y+z={ }^{12} C_3+{ }^{12} C_4+{ }^{13} C_3+{ }^{14} C_3+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$. Using the identity ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$: $x+y+z={ }^{13} C_4+{ }^{13} C_3+{ }^{14} C_3+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$. Continuing the application of the identity: $x+y+z={ }^{14} C_4+{ }^{14} C_3+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$. $x+y+z={ }^{15} C_4+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$. $x+y+z={ }^{16} C_4+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$. $x+y+z={ }^{17} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$. $x+y+z={ }^{18} C_5+{ }^{18} C_4+{ }^{19} C_4$. $x+y+z={ }^{19} C_5+{ }^{19} C_4={ }^{20} C_5$. Calculating the value: ${ }^{20} C_5 = \frac{20 	imes 19 	imes 18 	imes 17 	imes 16}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 19 	imes 3 	imes 17 	imes 16 = 19 	imes 17 	imes 48$.

If $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$,then $x+y+z=$

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