Download App

STD 11 - 6. permutation and combination — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 6. permutation and combination1 Mark
MCQ
$^{20}C_1 + 3 ^{20}C_2 + 3 ^{20}C_3 + ^{20}C_4$ is equal to-
  • A
    $^{20}C_4$
  • B
    $2. ^{21}C_4$
  • C
    $2. ^{22}C_4$
  • ✓
    $^{23}C_4$
✓

Answer

Correct option: D.
$^{23}C_4$
d
$({\,^{20}}{{\rm{C}}_1} + 2{\,^{20}}{{\rm{C}}_2} + {\,^{20}}{{\rm{C}}_3}{\rm{)}} + {{\rm{(}}^{20}}{{\rm{C}}_2} + 2{\,^{20}}{{\rm{C}}_3} + {\,^{20}}{{\rm{C}}_4}{\rm{)}}$

$ \Rightarrow {\,^{22}}{C_3} + {\,^{22}}{C_4} = {\,^{23}}{C_4}$

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

Explore more

More "M.C.Q (1 Marks)" from STD 11 - 6. permutation and combination→STD 11 - 6. permutation and combination — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Let $a_1, a_2, \ldots \ldots, a_n$ be in A.P. If $a_5=2 a_3$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.
→
Let $\mathrm{P}$ be a parabola with vertex $(2,3)$ and directrix $2 x+y=6$. Let an ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola $\mathrm{P}$. Then the square of the length of the latus rectum of $\mathrm{E}$, is
→
Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$ be the solution of the equation $4 x^4+8 x^3-17 x^2-12 x+9=0$ and $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$. Then the value of $\mathrm{m}$ is..........
→
The incorrect statement is
→
The distance between the foci of the ellipse $3{x^2} + 4{y^2} = 48$ is
→
If $x + \frac{1}{x} = 2\,\cos \theta ,$ then ${x^3} + \frac{1}{{{x^3}}} = $
→
The foci of the hyperbola $2{x^2} - 3{y^2} = 5$, is
→
If $A$ is the null set and $B$ is an infinite set, then what is $A \times B?$
→
Let $S={\theta \in\left(0, \frac{\pi}{2}\right): \sum_{m=1}^{9}}$

$\sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}$ Then.

→
The smallest positive integer $n$ for which $n! <\Big(\frac{\text{n+1}}{2}\Big)^\text{n}$ holds, is:
→