10 identical balls are distributed in 5 different boxes kept in a row and labeled $A,B,C,D$ and $E.$  Find the number of ways in which these balls can be distributed in the boxes if no two adjacent boxes remain empty.

Case –1:When no box remains empty it is equivalently distributing 10 coins in 5 beggar 
 $\underset{5}{\underbrace{00000}}\underset{4}{\underbrace{0000}}={}^9{C}_4=126$
Case –2: Exactly one is empty
 ${ }^5{C}_1\underset{5}{\underbrace{000000}}000=5.{}^9{C}_3=420$
Case –3: Exactly two remains empty
 ${(}^5{C}_2-{\text{\hspace{0.17em}two\hspace{0.17em}adjacent)\hspace{0.17em}\hspace{0.17em}}}^9{C}_2\underset{7}{\underbrace{0000000}}00(10-4)\times {}^9{C}_2$
 $6\times 36=216$
Case –4: Exactly three empty. There is only 1 way to select 3 if no two adjacent
Hence 1.  ${ }^9{C}_1=9000000000$
Total = 771 ways