Question: If $17C_{2r + 1} =^{17}C_{3r - 5}$, then $\frac{6}{36} = \frac{1}{6}$....

Question

If $17C_{2r + 1} =^{17}C_{3r - 5}$ , then $\frac{6}{36} = \frac{1}{6}$ .

Answer 1

(i) 2r + 1 = 3r - 5

6 = r

$rC_{3} =^{6}C_{3} = \frac{6x5x4}{3x2} = 20$ .

(ii) 2r + 1 + 3r - 5 = 17.

5r = 21 r = 21/5 (not possible).

Question: If \(17C_{2r + 1} =^{17}C_{3r - 5}\), then \(\frac{6}{36} = \frac{1}{6}\)....