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

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, $k+\lambda =7+15=22$