Solve the following

Question:

If 15Cr : 15Cr − 1 = 11 : 5, find r.

Solution:

Given:

 15Cr : 15Cr − 1 = 11 : 5

We have,

$\frac{{ }^{15} C_{r}}{{ }^{15} C_{r-1}}=\frac{11}{5}$

$\Rightarrow \frac{15-r+1}{r}=\frac{11}{5} \quad\left[\because \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\right]$

$\Rightarrow 75-5 r+5=11 r$

$\Rightarrow 16 r=80$

$\Rightarrow r=5$

 

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