Solve the following
Question:

If nC– 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. $\mathrm{e} \frac{r !(n-r)(n-r-1) !}{(r+1) r !(n-r-1) !}=\frac{3}{2}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.