Question

There are five different boxes and seven different balls. The number of ways in which these balls can be distributed so that box 2 and box 4 contain only 1 ball each and at least 1 box is empty is N. (Order of putting the balls in the boxes is NOT considered). Then the digit in the hundredth's place of N is ............

Answer 1

1 ball for box 2 & 1 for box 4 in ${}^7{C}_1{\times }^5{C}_1=42ways$

Remaining 5 balls are to be put in 3 boxes $=3\times (2^5-2)+3\times 1=93$

Total $=42\times 93=3906$