Discussion
In a Linear Programming Problem, suppose there are 3 basic variables and 2 non-basic variables, then the possible number of basic solutions are
Answer: C
A linear programming problem with 3 basic variables and 2 non-basic variables has 5 variables in all, so the number of constraints is 3, one for each basic variable. A basic solution is found by choosing 3 of the 5 variables as basic and setting the other two to zero, and the possible number of such choices is
\[ \binom{5}{3} = \frac{5!}{3!\,2!} = 10 \]The options 6, 8 and 12 do not equal this count. The keyed answer is 10.
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