Question

If $xϵN$, then range of function $f\left(x\right)=\frac{{}^{16-x}{C}_{x-1}}{{}^{20-3x}{C}_{x-5}}$ is

• A. $[1,15]$
• B. ${}^{14}{C}_4,\frac{{}^{10}{C}_5}{2}$
• C. ${}^{11}{C}_4$
• D. $[5,15]$

Answer 1

Answer: (B)

${}^{14}{C}_4,\frac{{}^{10}{C}_5}{2}$

In the function ${{}^{n}C}_r=\frac{n!}{r!(n-r)!},n,r>0\&n>r\therefore f\left(x\right)=\frac{{{}^{16-x}C}_{x-1}}{{{}^{20-3x}C}_{x-5}}\therefore 16-x<0x-1>020-3x>0x-5>016-x>x-120-3x>x-5$

Solving all these inequalities simultaneously we get,

$5\le x<7$

Since$x\in N\therefore x\in \{5,6\}isdomainoff\left(x\right)\therefore RangesetR\in f\left(5\right),f\left(6\right)f\left(5\right)=\frac{{{}^{16-5}C}_{5-1}}{{{}^{20-15}C}_0}=\frac{{{}^{11}C}_4}{{{}^{5}C}_0}=\frac{{{}^{11}C}_4}{\frac{5!}{0!5!}}={{}^{11}C}_{4}f\left(6\right)=\frac{{{}^{16-6}C}_{6-1}}{{{}^{20-3\times 6}C}_{6-5}}=\frac{{{}^{10}C}_5}{{{}^{2}C}_1}=\frac{{{}^{10}C}_5}{2}\therefore rangeis\{{{}^{11}C}_4,\frac{{{}^{10}C}_5}{2}\}$

Answer B