Question:
If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then r = ____________.
Solution:
$n C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$
here $\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{84}{36}$
i. e $\frac{n !}{r !(n-r) !} \times \frac{(r-1) !(n-r+1) !}{n !}=\frac{7}{3}$
i. e $\frac{(r-1) !(n-r+1)(n-r) !}{r(r-1) !(n-r) !}=\frac{7}{3}$
i. e $\frac{n-r+1}{r}=\frac{7}{3}$
i.e $3 n-3 r+3=7 r$
i. e $3 n+3=10 r$ ...$(1)$
also, $\frac{{ }^{n} C_{r+1}}{{ }^{n} C_{r}}=\frac{126}{84}$
$\frac{n !}{(r+1) !(n-r-1) !} \times \frac{r !(n-r) !}{n !}=\frac{3}{2}$
i. $\mathrm{e} \frac{r !(n-r)(n-r-1) !}{(r+1) r !(n-r-1) !}=\frac{3}{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.