MCQ
If ${^\text{20}}\text{C}_{3\text{r+1}}={^\text{20}}\text{C}_{\text{r-1}},$ is then r equal to:
  • 10
  • B
    11
  • C
    19
  • D
    12

Answer

Correct option: A.
10
$\text{r}+\text{1}+\text{r}-1=20$ $[\therefore \ ^\text{n}\text{C}_{\text{x}} = \ ^\text{n}\text{C}_{\text{y}} \Rightarrow \text{n} = \text{x} + \text{y} \text{ or } \text{x = y}]$
$\Rightarrow 2\text{r}=20$
$\Rightarrow \text{r}=10$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If the ${5^{th}}$ term of a $G.P.$ is $\frac{1}{3}$ and ${9^{th}}$ term is $\frac{{16}}{{243}}$, then the ${4^{th}}$ term will be
If $a \in R$ and the equation $ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$ (where $[x]$ denotes the greatest integer $\leq\,x$) has no integral solution, then all possible values of $a$ lie in the interval
Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$

Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$

Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$

One root of the following given equation $2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0$ is
If the lines $x + y = 6$ and $x + 2y = 4$ be diameters of the circle whose diameter is $20,$ then the equation of the circle is
If x - 1 > -x + 7 then which is true?
If one of the roots of the equation ${x^2} + ax + b = 0$ and ${x^2} + bx + a = 0$ is coincident, then the numerical value of $(a + b)$ is
Choose a number $n$ uniformly at random from the set $\{1,2, \ldots, 100\}$. Choose one of the first seven days of the year $2014$ at random and consider $n$ consecutive days starting from the chosen day. What is the probability that among the chosen $n$ days, the number of Sundays is different from the number of Mondays?
If $\frac{\pi }{2} < \alpha  < \frac{3}{2}\pi $ , then the modulus and argument of $(1 + cos\, 2\alpha ) + i\, sin\, 2\alpha $ is respectively
Probability is 0.45 that a dealer will sell at least 20 television sets during a day, and the probability is 0.74 that he will sell less that 24 televisions. The probability that he will sell 20, 21, 22 or 23 televisions during the day, is: