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3 votes

$56)x+y+z=11$ where $x\geq0,y\geq0,z\geq0$

using star and bars

number of  solutions$=\binom{11+3-1}{11}=\binom{13}{11}=\binom{13}{2}=78$

see this it is useful to read https://brilliant.org/wiki/integer-equations-star-and-bars/

$57)$ Referential integrity constraints works on the concept of the foreign key.

edited by
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1 votes
(x,y,z) combinations possible
(0,0,11) 3!/2! =3
(0,1,10) 3!=6
(0,2,9) 3!=6
(0,3,8) 3!=6
(0,4,7) 3!=6
(0,5,6) 3!=6
(1,9,1) 3!/2!=3
(1,8,2) 3!=6
(1,7,3) 3!=6
(1,6,4) 3!=6
(1,5,5) 3!=6
(2,7,2) 3!/2!=3
(2,6,3) 3!=6
(2,5,4) 3!=6
(3,5,3) 3!/2!=3

6*11 +3*4=66+12 =78

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