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Q.
If $^{n+1} C_{r+1} : {^{n}}C_{r}:{^{n-1}}{C_{r-1}}=11 : 6 :3,$ then $nr = $
Binomial Theorem
Solution:
Given,
$\frac{^{n+1}C_{r+1}}{^{n}C_{r}}=\frac{11}{6}$ or $\frac{ \frac{n+1}{r+1}\times^{n}C_{r}}{^{n}C_{r}}=\frac{11}{6}$
or $6n +6=11r +11$ or $6n -11r =5 ....\left(1\right)$
Also,
$\frac{^{n}C_{r}}{^{n-1}C_{r-1}}=\frac{6}{3}$ or $\frac{\frac{n}{r}\times^{n-1}C_{r-1}}{^{n-1}C_{r-1}}=\frac{6}{3}$ or $n = 2r ....\left(2\right)$
From (1) and (2), $r = 5$ and $n = 10$,
$\therefore nr=50$