MCQ
The value of ${}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{C_3}} $ is
  • A
    $^{56}{C_3}$
  • $^{56}{C_4}$
  • C
    $^{55}{C_4}$
  • D
    $^{55}{C_3}$

Answer

Correct option: B.
$^{56}{C_4}$
b
(b) ${\,^{50}}{C_4} + \,\left( {^{50}{C_3}{ + ^{51}}{C_3} + {\,^{52}}{C_3} + ......{\,^{55}}{C_3}} \right)$.

Taking first two terms together and adding them and following the same pattern, we get${\,^{56}}{C_4}$, $[As\,{\,^n}{C_r} + {\,^n}{C_{r - 1}} = {\,^{n + 1}}{C_r}]$.

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

The area of the triangle whose vertices are represented by the complex numbers $0, z,$ $z{e^{i\alpha }},$ $(0 < \alpha < \pi )$ equals
The number of words (with or without meaning) that can be formed from all the letters of the word $"LETTER"$ in which vowels never come together is
$\lim _{x \rightarrow 0} \frac{48}{x^4} \int \limits_0^x \frac{t^3}{t^6+1} d t$ is equal to $.......$.
Consider a function $f:R \to R$ 

$f\left( {x + a} \right) = \frac{1}{2} + \sqrt {f\left( x \right) - {f^2}\left( x \right)}$  a is a real constant, then $f(x)$ must be

Twenty tickets are marked the numbers $1, 2, ..... 20.$ If three tickets be drawn at random, then what is the probability that those marked $7$ and $11$ are among them
If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2}\,\, - \,ax\, + \,b}}{{x\, - \,1}}\,\, = \,3,$ then $a + b$ is equal to
If the three normals drawn to the parabola, $y ^{2}=2 x$ pass through the point $( a , 0) a \neq 0,$ then $'a'$ must be greater than
Let $X=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right], Y=\alpha l+\beta X+\gamma X^{2} \quad$ and $Z =\alpha^{2} I -\alpha \beta X +\left(\beta^{2}-\alpha \gamma\right) X ^{2}, \alpha, \beta, \gamma \in R$. If $Y ^{-1}=$ $\left[\begin{array}{ccc}\frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5}\end{array}\right]$, then $(\alpha-\beta+\gamma)^{2}$ is equal to
The roots of the equation ${x^4} - 8{x^2} - 9 = 0$ are
If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to ..........