Question

Range of $f\left(x\right){=}^{16-x}{C}_{2x-1}{+}^{20-3x}{C}_{4x-5}$ is

• A. $[728,1474]$
• B. $\{728,1617\}$
• C. $\{0,728\}$
• D. $\{728,1474\}$

Answer 1

Answer: (B)

$\{728,1617\}$

Given,

$f\left(x\right)={16-x_1}_{C}2x-1+{20-3x}_{C}4x-5$

$\begin{array}{cc}16-x⩾2x-1& \text{and}20-3x⩾4x-5\\ 17⩾3x& \text{and}25⩾7x\\ \frac{17}{3}⩾x-6& \text{and}\frac{25}{7}⩾x-\left(ii\right)\end{array}$

or.

$16-x>0,2x-1>0$ and $20-3x>0;4x-5>0$

$x<16;x>\frac{1}{2}$ and $x<\frac{20}{3};x>\frac{5}{4}$

After considering all the above cases, we get,

$1+\frac{1}{4}⩽x<3+\frac{4}{7}$

$\frac{5}{4}⩽x<\frac{25}{7}$ [x can be integer only]

$\therefore x=2,3$

Now, putting the value of $x=2$ in $f\left(x\right)$

$f\left(2\right)={}^4{c}_3+{}^{14}{c}_3=728$

$f\left(3\right)={}^{13}{c}_5+{}^{11}{c}_4=1287+330=1617$

Option B