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  • If <img alt="\mathrm{{ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84 \, and\, { }^{n} C_{r+1}=126}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%7B%20%7D%5E%7Bn%7D%20C_%7Br-1%7D%3D36%2C%7B

If \mathrm{{ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84 \, and\, { }^{n} C_{r+1}=126}, then the value of r is equal to

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

We have

\mathrm{\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{36}{84} \Rightarrow \frac{n !}{(r-1) !(n-r+1) !} \frac{r !(n-r) !}{n !}=\frac{3}{7}}

\mathrm{\Rightarrow \frac{r}{n-r+1}= \frac{3}{7}\Rightarrow 10r= 3n+3}

\mathrm{Similarly, \quad \frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}}=\frac{84}{126} \Rightarrow \frac{r+1}{n-r}=\frac{2}{3}}

\mathrm{\Rightarrow \quad 5 r+3=2 n \Rightarrow 5 r+3=2 n}

Solving we obtain \mathrm{r=3}.

Posted by

HARSH KANKARIA

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