If ${^{12} C}_{r + 1} = {^{12} C}_{3 r - 5}$ , find $r$ .

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Solution

Fact: If $^{n} C_{x} =^{n} C_{y}$ , then $x = y$ or $x + y = n$

Here $^{12} C_{r + 1} =^{12} C_{3 r - 5}$
$\Rightarrow r + 1 = 3 r - 5$ or $r + 1 + 3 r - 5 = 12$
$\Rightarrow 2 r = 6$ or $4 r = 16$
$\Rightarrow r = 3$ or $r = 4$

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