If 28C2r : 24C2r − 4 = 225 : 11, find r.
We have, ${ }^{28} C_{2 r}:{ }^{24} C_{2 r-4}=225: 11$
$\Rightarrow \frac{{ }^{28} C_{2 r}}{{ }^{24} C_{2 r-4}}=\frac{225}{11}$
$\Rightarrow \frac{28 !}{2 r !(28-2 r) !} \times \frac{(2 r-4) !(28-2 r) !}{24 !}=\frac{225}{11}$
$\Rightarrow \frac{28 \times 27 \times 26 \times 25}{2 r(2 r-1)(2 r-2)(2 r-3)}=\frac{225}{11}$
$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=\frac{28 \times 27 \times 26 \times 25 \times 11}{225}$
$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=28 \times 3 \times 26 \times 11$
$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=4 \times 7 \times 3 \times 13 \times 2 \times 11$
$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=(2 \times 7) \times 13 \times(3 \times 4) \times 11$
$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=14 \times 13 \times 12 \times 11$
$\Rightarrow 2 r=14$
$\Rightarrow r=7$
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