MHT CET · Maths · Binomial Theorem
The value of \(\frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{11} \mathrm{C}_{\mathrm{r}}}\), when both the numerator and denominator are at their greatest values, is
- A \(\frac{6}{11}\)
- B \(\frac{1}{11}\)
- C \(\frac{4}{11}\)
- D \(\frac{3}{11}\)
Answer & Solution
Correct Answer
(A) \(\frac{6}{11}\)
Step-by-step Solution
Detailed explanation
Greatest value of \({ }^{-10} \mathrm{C}_{\mathrm{r}}\) is at \(\mathrm{r}=5\) Greatest value of \({ }^{11} \mathrm{C}_{\mathrm{r}}\) is at \(\mathrm{r}=5\)
\(\therefore \frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{11} \mathrm{C}_{\mathrm{r}}}=\frac{{ }^{10} \mathrm{C}_5}{{ }^{11} \mathrm{C}_5}=\frac{\frac{10 !}{5 ! 5 !}}{\frac{11 !}{5 ! 6 !}}=\frac{6}{11}\)
\(\therefore \frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{11} \mathrm{C}_{\mathrm{r}}}=\frac{{ }^{10} \mathrm{C}_5}{{ }^{11} \mathrm{C}_5}=\frac{\frac{10 !}{5 ! 5 !}}{\frac{11 !}{5 ! 6 !}}=\frac{6}{11}\)
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