There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is

Question

There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is, 120, 89 . ( 8 ! ), 56, 42 × 5 4

MARKS App · Accepted Answer

We will use Gap method to solve this. Firstly, 7 White balls can be arranged in 7 ! ways. But Since all balls are identical so it should be 7 ! 7 ! ways. Now, A black ball can be between these white balls. So there are exactly 8 slots to place 3 black balls i.e., _ _ W _ _ W _ _ W _ _ W _ _ W _ _ W _ _ W _ _ This can be done in C 3 8 ways. Hence, the total number of ways is 7 ! 7 ! × C 3 = 56 8 ways.

Search any question & find its solution

Looking for more such questions to practice?