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  • If and <img alt="\mathrm{{ }^{x-1} C_{4}-{ }^{x-1} C_{3}<\frac{5}{4}\left({ }^{x

If \mathrm{x \in N} and \mathrm{{ }^{x-1} C_{4}-{ }^{x-1} C_{3}<\frac{5}{4}\left({ }^{x-2} P_{2}\right)}, then value of \mathrm{x} can take

Option: 1

8 values


Option: 2

7 values


Option: 3

 6 values


Option: 4

 none of these


Answers (1)

best_answer

For  1 \leq x \leq 3$, each of ${ }^{x-1} C_{4},{ }^{x-1} C_{3}$ and ${ }^{x-2} P_{2} is zero. For x=4,{ }^{x-1} C_{4}=0,{ }^{x-1} C_{3}=1$ and ${ }^{x-2} P_{2}=2 and the inequality is valid trivally. For x \geq 5, we have

\mathrm{ \frac{{ }^{x-1} C_{4}}{{ }^{x-2} P_{2}}-\frac{{ }^{x-1} C_{3}}{{ }^{x-2} P_{2}}<\frac{5}{4} }
\mathrm{ \Rightarrow \quad \frac{(x-1)(x-4)}{4 !}-\frac{x-1}{3 !}<\frac{5}{4} }
\mathrm{ \Rightarrow \quad x^{2}-9 x-22<0 \Rightarrow(x-11)(x+2)<0 }
\mathrm{ \Rightarrow \quad x-11<0[\because x+2>0]}

Thus, \mathrm{ 5 \leq x<11}.
Hence , \mathrm{ x} can take 7 values viz. 4,5,6, 7,8,9 and 10 .
 

Posted by

rishi.raj

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