Question
If $^n{P_r}$=$ 720$.$^n{C_r},$ then $r$ is equal to

Answer

a
(a) $^n{P_r}\, = 720.{\,^n}{C_r}$

==> $^n{P_r}\, \div {\,^n}{C_r}\, = 720$

==> $\,r\,! = 720 = 6\,!$

$==> r = 6.$

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 lines $L_1, L_2, \ldots, I_{20}$ are distinct. For $n =1,2,3, \ldots ., 10$ all the lines $L _{2 n -1}$ are parallel to each other and all the lines $L _{2 n}$ pass through a given point $P$. The maximum number of points of intersection of pairs of lines from the set $\left\{L_1, L_2, \ldots, L_{20}\right\}$ is equal to :
Two tangents are drawn from the point $\mathrm{P}(-1,1)$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-6 \mathrm{y}+6=0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $A B$ and $A D$ are equal, then the area of the triangle $A B D$ is eqaul to:
If the first term of an $A.P. $ be $10$, last term is $50$ and the sum of all the terms is $300$, then the number of terms are
If $a, b, c$ are in $GP$ and $4a, 5b, 4c$ are in $AP$ such that $a + b + c = 70$,  then value of $a^3 + b^3 + c^3$ is
In a plane there are $10$ points out of which $4$ are collinear, then the number of triangles that can be formed by joining these points are
Let $a, b, c$ be the length of three sides of a triangle satisfying the condition $\left(a^2+b^2\right) x^2-2 b(a+c)$. $x+\left(b^2+c^2\right)=0$. If the set of all possible values of $x$ is the interval $(\alpha, \beta)$, then $12\left(\alpha^2+\beta^2\right)$ is equal to $..........$
The derivative of $f(x) = \,|{x^2} - x|$ at  $x = 2$  is
If ${\cot ^{ - 1}}\frac{n}{\pi } > \frac{\pi }{6},\,\,n \in N$ , then the maximum  value of $n$ is
Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals
The number of triples $(x, y, z)$ of real numbers satisfying the equation $x^4+y^4+z^4+1=4 x y z$ is