Discussion
How many solutions do the following equation have
\(x_1 + x_2 + x_3 = 11\)
where \(x_1 \ge 1, x_2 \ge 2, x_3 \ge 3\)
Answer: D
To count the solutions of \(x_1 + x_2 + x_3 = 11\) with \(x_1 \ge 1\), \(x_2 \ge 2\) and \(x_3 \ge 3\), substitute \(y_1 = x_1 - 1\), \(y_2 = x_2 - 2\) and \(y_3 = x_3 - 3\), which are non-negative. Then
\[ y_1 + y_2 + y_3 = 11 - 6 = 5 \]The number of non-negative solutions of this equation in 3 variables is
\[ C(5 + 3 - 1, 5) = C(7, 5) = 21 \]The option C(11, 3) = 165 and C(14, 11) = 364 are too large, and C(7, 11) has no meaning. The keyed answer is C(7, 5).
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