In how many ways can 21 white balls and 19 bl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To arrange $21$ white balls and $19$ black balls such that no two black balls are together,we use the gap method.First,arrange the $21$ white ball

Vedclass · Accepted Answer

$1540$

Vedclass · Answer

(B) To arrange $21$ white balls and $19$ black balls such that no two black balls are together,we use the gap method.First,arrange the $21$ white balls in a row. This creates $22$ possible gaps (including the ends) where the black balls can be placed.Number of ways to arrange $21$ white balls $= 1$.Number of ways to choose $19$ gaps out of $22$ for the black balls $= ^{22}C_{19}$.Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we have $^{22}C_{19} = ^{22}C_{3}$.$^{22}C_{3} = \frac{22 	imes 21 	imes 20}{3 	imes 2 	imes 1} = 22 	imes 7 	imes 10 = 1540$.Total arrangements $= 1540 	imes 1 = 1540$.

In how many ways can $21$ white balls and $19$ black balls be arranged in a row such that no two black balls are together?

Similar Questions

There are $8$ points in a plane,out of which $4$ are collinear. How many triangles can be formed by joining these points?

Let $P$ be a closed polygon with $10$ sides and $10$ vertices (assume that the sides do not intersect except at the vertices). Let $k$ be the number of interior angles of $P$ that are greater than $180^{\circ}$. The maximum possible value of $k$ is

Three out of $6$ vertices of a regular hexagon are chosen at random. The probability that the triangle formed with these three vertices is an equilateral triangle is

The number of diagonals in a convex polygon with $n$ sides is .....

Let $t_n$ denote the number of integral-sided triangles with distinct sides chosen from $\{1, 2, 3, \ldots, n\}$. Then,$t_{20} - t_{19}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module