Question

Seven white balls and three black balls are randomly placed in a row. If the probability that no two black balls are placed adjacently equals $\frac{k}{\lambda }$ , then the value $k+\lambda$ is

Answer 1

Answer: 22

Total number of ways to arrange seven white balls and three black balls in a row $=\frac{120\left(10!\right)}{7!3!}$
There are $6$ blank spaces between $7$ white balls. And also there is one space before the first ball and another space after the last ball. Hence, the total number of blank spaces is $8$ .
So, $3$ black balls are arranged in $8$ places, such that no two black balls are together $=\_{}^8{C}_3^{}=56$
The required probability $=\frac{56}{120}=\frac{7}{15}$
Therefore, $k=7,\lambda =15$
Hence, $k+\lambda =7+15=22$