what are the number of solutions for x+y+z=12, if x,y & z can take values from 1 to 7 ?
Choose the most appropriate option: A) 34 B) 35 C) 36 D) 37 E) 38
if x=1, (y,z) can be (4,7)(5,6)(6,5)(7,4)   if x=2, (y,z) can be (3,7)(4,6)(5,5)(6,4)(7,3)   if x=3, (y,z) can be (2,7)(3,6)(4,5)(5,4)(6,3)(7,2)   if x=4, (y,z) can be (1,7)(2,6)(3,5)(4,4)(5,3)(6,2)(7,1)   if x=5, (y,z) can be (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)   if x=6, (y,z) can be (1,5)(2,4)(3,3)(4,2)(5,1)   if x=7, (y,z) can be (1,4)(2,3)(3,2)(4,1)   Total Combination = 4+5+6+7+6+5+4= 
<gadget id="calculator">4+5+6+7+6+5+4</gadget>
<output>37</output>
37   ANSWER:37
<result>D</result>
what are the number of solutions for x+y+z=12, if x,y & z can take values from 1 to 7 ?