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If 28C2r : 24C2r − 4 = 225 : 11, find r.

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Question

If 28C2r : 24C2r − 4 = 225 : 11, find r.

Sum
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Solution

We have,  28C2r : 24C2r − 4 = 225 : 11

\[\Rightarrow \frac{{}^{28} C_{2r}}{{}^{24} C_{2r - 4}} = \frac{225}{11}\]
\[ \Rightarrow \frac{28!}{2r! (28 - 2r)!} \times \frac{(2r - 4)! (28 - 2r)!}{24!} = \frac{225}{11}\]
\[ \Rightarrow \frac{28 \times 27 \times 26 \times 25}{2r (2r - 1) (2r - 2) (2r - 3)} = \frac{225}{11}\]
\[ \Rightarrow 2r (2r - 1) (2r - 2) (2r - 3) = \frac{28 \times 27 \times 26 \times 25 \times 11}{225}\]
\[ \Rightarrow 2r (2r - 1) (2r - 2) (2r - 3) = 28 \times 3 \times 26 \times 11\]
\[ \Rightarrow 2r (2r - 1) (2r - 2) (2r - 3) = 4 \times 7 \times 3 \times 13 \times 2 \times 11\]
\[ \Rightarrow 2r (2r - 1) (2r - 2) (2r - 3) = (2 \times 7) \times 13 \times (3 \times 4) \times 11\]
\[ \Rightarrow 2r (2r - 1) (2r - 2) (2r - 3) = 14 \times 13 \times 12 \times 11\]
\[ \Rightarrow 2r = 14\]
\[ \Rightarrow r = 7\]
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Chapter 17: Combinations - Exercise 17.1 [Page 8]

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R.D. Sharma Mathematics [English] Class 11
Chapter 17 Combinations
Exercise 17.1 | Q 11 | Page 8

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