Question

. If n cr–1 = 56, ncr

= 28 and n cr+1 = 8, then r is equal to​

Answer 1

Answer:

r=6

Step-by-step explanation:

Check the Photo Attached !

Answer 2

Answer :

r = 6

Given :

$^nC_{r-1}=56 \\ ^nC_r=28\\^nC_{r+1}=8$

To find :

$r= \ ?$

Solution :

      $=> \frac{^nC_{r-1}}{^nC_r}=\frac{56}{28} =2 \\\\ => \frac{n!r!(n-r)!}{(r-1)!(n-(r-1))!n!} =2 \\\\ => \frac{r}{n-r+1} =2 \\\\ => r=2n-2r+2\\\\ =>3r=2n+2 \\\\ =>r=\frac{2n+2}{3} \\\\ ----------\\\\ =>\frac{^nC_r}{^nC_{r+1}} =\frac{28}{8} =\frac{7}{2} \\\\ =>\frac{n!(n-(r+1))!(r+1)!}{n!r!(n-r)!} =\frac{7}{2} \\\\ => \frac{r+1}{n-r} =\frac{7}{2} \\\\ =>2r+2=7n-7r \\\\ => 9r=7n-2 \\\\ => 9(\frac{2n+2}{3} )=7n-2 \\\\ =>3(2n+2)=7n-2 \\\\ =>6n+6=7n-2 \\\\ =>7n-6n=6+2 \\\\ =>\textbf{n = 8}$

     

       $\implies r=\frac{2n+2}{3} \\\\ \implies r=\frac{2(8)+2}{3} \\\\ \implies r=\frac{18}{3} \\\\ \implies \textbf{r = 6}$