CAT Exam Model Paper 5 with solutions for free online practice

Case 1:
All four balls in the same box: 1 way
Since all the boxes are identical.
Case 2:
Two balls in 1 box and the remaining two in another box:
$\frac{{}_{}^4{\mathit{C}}_2^{}}{2!}=3$ ways
It is divided by 2! because the boxes are identical.
Case 3:
Two balls in a box and one ball in each of the remaining two boxes:
${}_{}^4{\mathit{C}}_2^{}=6$ ways
Note: We need not select a box for the remaining two balls as they will go one each in the remaining two boxes in one way only.
Case 4:
Three balls in one box and one remaining ball in another box:
${}_{}^4{\mathit{C}}_3^{}=4$ ways
Since there is only one way of selecting the other box as the boxes are identical.
Total ways = 1 + 3 + 6 + 4 = 14